Chapter 8: Introduction to Trigonometry

Ans: For a right triangle we have an identity ${{\operatorname{cosec}}^{2}}A=1+{{\cot }^{2}}A$.

Let us consider the above identity, we get

${{\operatorname{cosec}}^{2}}A=1+{{\cot }^{2}}A$

Now, reciprocating both sides we get

$\Rightarrow \dfrac{1}{{{\operatorname{cosec}}^{2}}A}=\dfrac{1}{1+{{\cot }^{2}}A}$

Now, we know that $\dfrac{1}{{{\operatorname{cosec}}^{2}}A}={{\sin }^{2}}A$, we get

$\Rightarrow {{\sin }^{2}}A=\dfrac{1}{1+{{\cot }^{2}}A}$

$\Rightarrow \sin A=\pm \dfrac{1}{\sqrt{1+{{\cot }^{2}}A}}$

Now, we know that sine value will be negative for angles greater than $180{}^\circ $, for a triangle sine value is always positive with respect to an angle. Then we will consider only positive value.

$\therefore \sin A=\dfrac{1}{\sqrt{1+{{\cot }^{2}}A}}$ 

We know that $\tan A=\dfrac{1}{\cot A}$ 

Also, we will use the identity ${{\sec }^{2}}A=1+{{\tan }^{2}}A$, we get

${{\sec }^{2}}A=1+{{\tan }^{2}}A$

$\Rightarrow {{\sec }^{2}}A=1+\dfrac{1}{{{\cot }^{2}}A}$

$\Rightarrow {{\sec }^{2}}A=\dfrac{{{\cot }^{2}}A+1}{{{\cot }^{2}}A}$

\[\Rightarrow \sec A=\dfrac{\sqrt{{{\cot }^{2}}A+1}}{\sqrt{{{\cot }^{2}}A}}\]

\[\therefore \sec A=\dfrac{\sqrt{{{\cot }^{2}}A+1}}{\cot A}\]

Ans: 

We know that $\cos A=\dfrac{1}{\sec A}$.

$\therefore \cos A=\dfrac{1}{\sec A}$

For a right triangle we have an identity ${{\sin }^{2}}A+{{\cos }^{2}}A=1$.

Let us consider the above identity, we get

${{\sin }^{2}}A+{{\cos }^{2}}A=1$

Now, we know that $\cos A=\dfrac{1}{\sec A}$, we get

$\Rightarrow {{\sin }^{2}}A=1-{{\cos }^{2}}A$

$\Rightarrow {{\sin }^{2}}A=1-\dfrac{1}{{{\sec }^{2}}A}$

$\Rightarrow \sin A=\sqrt{1-{{\left( \dfrac{1}{\sec A} \right)}^{2}}}$

$\therefore \sin A=\dfrac{\sqrt{{{\sec }^{2}}A-1}}{\sec A}$

Also, we will use the identity ${{\sec }^{2}}A=1+{{\tan }^{2}}A$, we get

${{\tan }^{2}}A={{\sec }^{2}}A-1$

$\therefore \tan A=\sqrt{{{\sec }^{2}}A-1}$ 

Now, we know that $\cot A=\dfrac{\cos A}{\sin A}$, we get

$\Rightarrow \cot A=\dfrac{\dfrac{1}{\sec A}}{\dfrac{\sqrt{{{\sec }^{2}}A-1}}{\sec A}}$

$\therefore \cot A=\dfrac{1}{\sqrt{{{\sec }^{2}}A-1}}$

We know that $cosecA=\dfrac{1}{\sin A}$, we get $\therefore cosecA=\dfrac{\sec A}{\sqrt{{{\sec }^{2}}A-1}}$

Ans: The given expression is $9{{\sec }^{2}}A-9{{\tan }^{2}}A$.

The given expression can be written as 

$\Rightarrow 9{{\sec }^{2}}A-9{{\tan }^{2}}A=9\left( {{\sec }^{2}}A-{{\tan }^{2}}A \right)$

Now, we will use the identity ${{\sec }^{2}}A=1+{{\tan }^{2}}A$, we get

${{\sec }^{2}}A-{{\tan }^{2}}A=1$

$\Rightarrow 9{{\sec }^{2}}A-9{{\tan }^{2}}A=9\left( 1 \right)$

$\therefore 9{{\sec }^{2}}A-9{{\tan }^{2}}A=9$

Therefore, option (B) is the correct answer.

1. $\left( 1+\tan \theta +\sec \theta  \right)\left( 1+\cot \theta -\operatorname{cosec}\theta  \right)$ 

1. $0$

2. $1$

3. $2$

4. $-1$

Ans: The given expression is $\left( 1+\tan \theta +\sec \theta  \right)\left( 1+\cot \theta -\operatorname{cosec}\theta  \right)$.

We know that the trigonometric functions have values as:

$\tan \theta =\dfrac{\sin \theta }{\cos \theta }$, $\cot \theta =\dfrac{\cos \theta }{\sin \theta }=\dfrac{1}{\tan \theta }$, $cosec\theta =\dfrac{1}{\sin \theta }$ and $sec\theta =\dfrac{1}{\cos \theta }$

Substituting these values in the given expression, we get

$\Rightarrow \left( 1+\tan \theta +\sec \theta  \right)\left( 1+\cot \theta -\operatorname{cosec}\theta  \right)=\left( 1+\dfrac{\sin \theta }{\cos \theta }+\dfrac{1}{\cos \theta } \right)\left( 1+\dfrac{\cos \theta }{\sin \theta }-\dfrac{1}{\sin \theta } \right)$

$\Rightarrow \left( 1+\tan \theta +\sec \theta  \right)\left( 1+\cot \theta -\operatorname{cosec}\theta  \right)=\left( \dfrac{\cos \theta +\sin \theta +1}{\cos \theta } \right)\left( \dfrac{\sin \theta +\cos \theta -1}{\sin \theta } \right)$

Now, by applying the identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$, we get

$\Rightarrow \left( 1+\tan \theta +\sec \theta  \right)\left( 1+\cot \theta -\operatorname{cosec}\theta  \right)=\dfrac{{{\left( \sin \theta +\cos \theta  \right)}^{2}}-{{1}^{2}}}{\sin \theta \cos \theta }$

$\Rightarrow \left( 1+\tan \theta +\sec \theta  \right)\left( 1+\cot \theta -\operatorname{cosec}\theta  \right)=\dfrac{{{\sin }^{2}}\theta +{{\cos }^{2}}\theta +2\sin \theta \cos \theta -1}{\sin \theta \cos \theta }$

Now, by applying the trigonometric identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, we get

$\Rightarrow \left( 1+\tan \theta +\sec \theta  \right)\left( 1+\cot \theta -\operatorname{cosec}\theta  \right)=\dfrac{1+2\sin \theta \cos \theta -1}{\sin \theta \cos \theta }$

$\Rightarrow \left( 1+\tan \theta +\sec \theta  \right)\left( 1+\cot \theta -\operatorname{cosec}\theta  \right)=\dfrac{2\sin \theta \cos \theta }{\sin \theta \cos \theta }$

$\therefore \left( 1+\tan \theta +\sec \theta  \right)\left( 1+\cot \theta -\operatorname{cosec}\theta  \right)=2$

Therefore, option (C) is the correct answer.

1. $\left( \sec A+\tan A \right)\left( 1-\sin A \right)=$ ………

1. $\sec A$ 

2. $\sin A$ 

3. $cosecA$ 

4. $\cos A$ 

Ans: Given expression is $\left( \sec A+\tan A \right)\left( 1-\sin A \right)$.

We know that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and $sec\theta =\dfrac{1}{\cos \theta }$

Substituting these values in the given expression, we get

$\left( \sec A+\tan A \right)\left( 1-\sin A \right)=\left( \dfrac{1}{\cos A}+\dfrac{\sin A}{\cos A} \right)\left( 1-\sin A \right)$

$\Rightarrow \left( \sec A+\tan A \right)\left( 1-\sin A \right)=\left( \dfrac{1+\sin A}{\cos A} \right)\left( 1-\sin A \right)$

$\Rightarrow \left( \sec A+\tan A \right)\left( 1-\sin A \right)=\left( \dfrac{\left( 1+\sin A \right)\left( 1-\sin A \right)}{\cos A} \right)$

Now, by applying the identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$, we get

$\Rightarrow \left( \sec A+\tan A \right)\left( 1-\sin A \right)=\left( \dfrac{{{1}^{2}}-{{\sin }^{2}}A}{\cos A} \right)$

Now, we know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, we get

$\Rightarrow \left( \sec A+\tan A \right)\left( 1-\sin A \right)=\left( \dfrac{{{\cos }^{2}}A}{\cos A} \right)$

$\therefore \left( \sec A+\tan A \right)\left( 1-\sin A \right)=\cos A$

Therefore, option (D) is the correct answer.

1. $\dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}$ 

1. ${{\sec }^{2}}A$ 

2. $-1$ 

3. ${{\cot }^{2}}A$ 

4. ${{\tan }^{2}}A$ 

Ans: Given expression is $\dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}$.

We know that the trigonometric functions have values as:

$\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and $\cot \theta =\dfrac{\cos \theta }{\sin \theta }=\dfrac{1}{\tan \theta }$.

Substituting these values in the given expression, we get

$\dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}=\dfrac{1+\dfrac{{{\sin }^{2}}A}{{{\cos }^{2}}A}}{1+\dfrac{{{\cos }^{2}}A}{{{\sin }^{2}}A}}$

$\Rightarrow \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}=\dfrac{\dfrac{{{\cos }^{2}}A+{{\sin }^{2}}A}{{{\cos }^{2}}A}}{\dfrac{{{\sin }^{2}}A+{{\cos }^{2}}A}{{{\sin }^{2}}A}}$

Now, we know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, we get

$\Rightarrow \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}=\dfrac{\dfrac{1}{{{\cos }^{2}}A}}{\dfrac{1}{{{\sin }^{2}}A}}$

$\Rightarrow \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}=\dfrac{{{\sin }^{2}}A}{{{\cos }^{2}}A}$

$\Rightarrow \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}={{\tan }^{2}}A$

Therefore, option (D) is the correct answer.

Ans: Given expression is ${{\left( cosec\theta -\cot \theta  \right)}^{2}}=\dfrac{1-\cos \theta }{1+\cos \theta }$.

Let us consider the LHS of the given expression, we get

$LHS={{\left( cosec\theta -\cot \theta  \right)}^{2}}$

Now, we know that $\cot \theta =\dfrac{\cos \theta }{\sin \theta }=\dfrac{1}{\tan \theta }$ and $cosec\theta =\dfrac{1}{\sin \theta }$.

By substituting the values, we get

$\Rightarrow {{\left( cosec\theta -\cot \theta  \right)}^{2}}={{\left( \dfrac{1}{\sin \theta }-\dfrac{\cos \theta }{\sin \theta } \right)}^{2}}$

$\Rightarrow {{\left( cosec\theta -\cot \theta  \right)}^{2}}={{\left( \dfrac{1-\cos \theta }{\sin \theta } \right)}^{2}}$

$\Rightarrow {{\left( cosec\theta -\cot \theta  \right)}^{2}}=\dfrac{{{\left( 1-\cos \theta  \right)}^{2}}}{{{\sin }^{2}}\theta }$

Now, we know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, we get

$\Rightarrow {{\left( cosec\theta -\cot \theta  \right)}^{2}}=\dfrac{{{\left( 1-\cos \theta  \right)}^{2}}}{1-{{\cos }^{2}}\theta }$

Now, by applying the identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$, we get

$\Rightarrow {{\left( cosec\theta -\cot \theta  \right)}^{2}}=\dfrac{{{\left( 1-\cos \theta  \right)}^{2}}}{\left( 1-\cos \theta  \right)\left( 1+\cos \theta  \right)}$

$\Rightarrow {{\left( cosec\theta -\cot \theta  \right)}^{2}}=\dfrac{\left( 1-\cos \theta  \right)}{\left( 1+\cos \theta  \right)}$

$\Rightarrow {{\left( cosec\theta -\cot \theta  \right)}^{2}}=RHS$

$\therefore {{\left( cosec\theta -\cot \theta  \right)}^{2}}=\dfrac{1-\cos \theta }{1+\cos \theta }$

Hence proved

1. $\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=2\sec A$

Ans: Given expression is $\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=2\sec A$.

Let us consider the LHS of the given expression, we get

$LHS=\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}$

Now, taking LCM, we get

$\Rightarrow \dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=\dfrac{{{\cos }^{2}}A+\left( 1+\sin A \right)\left( 1+\sin A \right)}{\left( 1+\sin A \right)\cos A}$

$\Rightarrow \dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=\dfrac{{{\cos }^{2}}A+{{\sin }^{2}}A+2\sin A+1}{\left( 1+\sin A \right)\cos A}$

Now, we know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, we get

$\Rightarrow \dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=\dfrac{1+2\sin A+1}{\left( 1+\sin A \right)\cos A}$

$\Rightarrow \dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=\dfrac{2+2\sin A}{\left( 1+\sin A \right)\cos A}$

$\Rightarrow \dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=\dfrac{2\left( 1+\sin A \right)}{\left( 1+\sin A \right)\cos A}$

$\Rightarrow \dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=\dfrac{2}{\cos A}$

We know that $sec\theta =\dfrac{1}{\cos \theta }$, we get

$\Rightarrow \dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=2\sec A$

\[\Rightarrow \dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=RHS\]

\[\therefore \dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=2\sec A\]

Hence proved

1. $\dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=1+\sec \theta cosec\theta $ 

Ans: Given expression is $\dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=1+\sec \theta \operatorname{cosec}\theta $.

Let us consider the LHS of the given expression, we get

$LHS=\dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }$

Now, we know that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and $\cot \theta =\dfrac{\cos \theta }{\sin \theta }=\dfrac{1}{\tan \theta }$.

By substituting the values, we get

\[\Rightarrow \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=\left( \dfrac{\dfrac{\sin \theta }{\cos \theta }}{1-\dfrac{\cos \theta }{\sin \theta }}+\dfrac{\dfrac{\cos \theta }{\sin \theta }}{1-\dfrac{\sin \theta }{\cos \theta }} \right)\]

\[\Rightarrow \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=\left( \dfrac{\dfrac{\sin \theta }{\cos \theta }}{\dfrac{\sin \theta -\cos \theta }{\sin \theta }}+\dfrac{\dfrac{\cos \theta }{\sin \theta }}{\dfrac{\cos \theta -\sin \theta }{\cos \theta }} \right)\]

 \[\Rightarrow \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=\left( \dfrac{{{\sin }^{2}}\theta }{\cos \theta \left( \sin \theta -\cos \theta  \right)}+\dfrac{{{\cos }^{2}}\theta }{\sin \theta \left( \sin \theta -\cos \theta  \right)} \right)\]

\[\Rightarrow \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=\dfrac{1}{\left( \sin \theta -\cos \theta  \right)}\left( \dfrac{{{\sin }^{2}}\theta }{\cos \theta }+\dfrac{{{\cos }^{2}}\theta }{\sin \theta } \right)\]

\[\Rightarrow \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=\dfrac{1}{\left( \sin \theta -\cos \theta  \right)}\left( \dfrac{{{\sin }^{3}}\theta -{{\cos }^{3}}\theta }{\sin \theta \cos \theta } \right)\]

Now, by applying the identity \[{{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)\], we get

\[\Rightarrow \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=\dfrac{1}{\left( \sin \theta -\cos \theta  \right)}\left[ \dfrac{\left( \sin \theta -\cos \theta  \right)\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta +\sin \theta \cos \theta  \right)}{\sin \theta \cos \theta } \right]\]

Now, we know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, we get

\[\Rightarrow \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=\dfrac{1}{\left( \sin \theta -\cos \theta  \right)}\left[ \dfrac{\left( \sin \theta -\cos \theta  \right)\left( 1+\sin \theta \cos \theta  \right)}{\sin \theta \cos \theta } \right]\]

\[\Rightarrow \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=\dfrac{\left( 1+\sin \theta \cos \theta  \right)}{\sin \theta \cos \theta }\]

\[\Rightarrow \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=\dfrac{1}{\sin \theta \cos \theta }+\dfrac{\sin \theta \cos \theta }{\sin \theta \cos \theta }\]

\[\Rightarrow \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=\dfrac{1}{\sin \theta \cos \theta }+1\]

We know that $cosec\theta =\dfrac{1}{\sin \theta }$ and $sec\theta =\dfrac{1}{\cos \theta }$, we get

\[\Rightarrow \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=\sec \theta cosec\theta +1\]

\[\Rightarrow \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=1+\sec \theta cosec\theta \]

\[\Rightarrow \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=RHS\]

\[\therefore \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=1+\sec \theta cosec\theta \]

Hence proved

1. $\dfrac{1+\sec A}{\sec A}=\dfrac{{{\sin }^{2}}A}{1-\cos A}$ 

Ans: Given expression is $\dfrac{1+\sec A}{\sec A}=\dfrac{{{\sin }^{2}}A}{1-\cos A}$.

Let us consider the LHS of the given expression, we get

$LHS=\dfrac{1+\sec A}{\sec A}$

Now, we know that $\sec \theta =\dfrac{1}{\cos \theta }$.

By substituting the value, we get

\[\Rightarrow \dfrac{1+\sec A}{\sec A}=\dfrac{1+\dfrac{1}{\cos A}}{\dfrac{1}{\cos A}}\]

\[\Rightarrow \dfrac{1+\sec A}{\sec A}=\dfrac{\dfrac{\cos A+1}{\cos A}}{\dfrac{1}{\cos A}}\]

 \[\Rightarrow \dfrac{1+\sec A}{\sec A}=\cos A+1\]

Multiply and divide by $\left( 1-\cos A \right)$, we get

 \[\Rightarrow \dfrac{1+\sec A}{\sec A}=\dfrac{\left( 1+\cos A \right)\left( 1-\cos A \right)}{\left( 1-\cos A \right)}\]

Now, by applying the identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$, we get

\[\Rightarrow \dfrac{1+\sec A}{\sec A}=\dfrac{1-{{\cos }^{2}}A}{\left( 1-\cos A \right)}\]

Now, we know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, we get

\[\Rightarrow \dfrac{1+\sec A}{\sec A}=\dfrac{{{\sin }^{2}}A}{\left( 1-\cos A \right)}\]

\[\Rightarrow \dfrac{1+\sec A}{\sec A}=RHS\]

\[\therefore \dfrac{1+\sec A}{\sec A}=\dfrac{{{\sin }^{2}}A}{1-\cos A}\]

Hence proved

1. $\dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=cosecA+\cot A$ 

Ans: Given expression is $\dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=cosecA+\cot A$.

Now, let us consider the LHS of the given expression, we get

$LHS=\dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}$

Dividing numerator and denominator by $\sin A$, we get

$\Rightarrow \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\dfrac{\dfrac{\cos A}{\sin A}-\dfrac{\sin A}{\sin A}+\dfrac{1}{\sin A}}{\dfrac{\cos A}{\sin A}+\dfrac{\sin A}{\sin A}-\dfrac{1}{\sin A}}$

Now, we know that $\cot \theta =\dfrac{\cos \theta }{\sin \theta }=\dfrac{1}{\tan \theta }$ and $cosec\theta =\dfrac{1}{\sin \theta }$, we get

$\Rightarrow \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\dfrac{\cot A-1+\operatorname{cosec}A}{\cot A+1-\operatorname{cosec}A}$

Now, by applying the identity ${{\operatorname{cosec}}^{2}}A=1+{{\cot }^{2}}A$, substitute $1={{\cot }^{2}}A-{{\operatorname{cosec}}^{2}}A$, we get

$\Rightarrow \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\dfrac{\cot A-\left( {{\cot }^{2}}A-{{\operatorname{cosec}}^{2}}A \right)+\operatorname{cosec}A}{\cot A+{{\cot }^{2}}A-{{\operatorname{cosec}}^{2}}A-\operatorname{cosec}A}$

$\Rightarrow \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\dfrac{\cot A-{{\cot }^{2}}A+{{\operatorname{cosec}}^{2}}A+\operatorname{cosec}A}{\cot A+{{\cot }^{2}}A-{{\operatorname{cosec}}^{2}}A-\operatorname{cosec}A}$

$\Rightarrow \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\dfrac{{{\left( \cot A-1+\operatorname{cosec}A \right)}^{2}}}{{{\cot }^{2}}A-1+{{\operatorname{cosec}}^{2}}A+2\operatorname{cosec}A}$

$\Rightarrow \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\dfrac{{{\left( \cot A-1+\operatorname{cosec}A \right)}^{2}}}{{{\cot }^{2}}A-1+{{\operatorname{cosec}}^{2}}A+2\operatorname{cosec}A}$

$\Rightarrow \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\dfrac{2{{\operatorname{cosec}}^{2}}A+2\cot A\operatorname{cosec}A-2\cot A-2\operatorname{cosec}A}{{{\cot }^{2}}A-1+{{\operatorname{cosec}}^{2}}A+2\operatorname{cosec}A}$

$\Rightarrow \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\dfrac{2\operatorname{cosec}A\left( \cot A-\operatorname{cosec}A \right)-2\left( \cot A-\operatorname{cosec}A \right)}{{{\cot }^{2}}A-1+{{\operatorname{cosec}}^{2}}A+2\operatorname{cosec}A}$

$\Rightarrow \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\dfrac{\left( 2\operatorname{cosec}A-2 \right)\left( \cot A-\operatorname{cosec}A \right)}{1-1+2\operatorname{cosec}A}$

$\Rightarrow \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\dfrac{\left( 2\operatorname{cosec}A-2 \right)\left( \cot A-\operatorname{cosec}A \right)}{2\operatorname{cosec}A}$

$\Rightarrow \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\operatorname{cosec}A+\cot A$

$\Rightarrow \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=RHS$

$\therefore \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\operatorname{cosec}A+\cot A$ 

Hence proved

1. $\sqrt{\dfrac{1+\sin A}{1-\sin A}}=\sec A+\tan A$ 

Ans: Given expression is $\sqrt{\dfrac{1+\sin A}{1-\sin A}}=\sec A+\tan A$.

Let us consider the LHS of the given expression, we get

$LHS=\sqrt{\dfrac{1+\sin A}{1-\sin A}}$

Now, multiply and divide the expression by $\sqrt{1+\sin A}$, we get

$\Rightarrow \sqrt{\dfrac{1+\sin A}{1-\sin A}}=\sqrt{\dfrac{\left( 1+\sin A \right)\left( 1+\sin A \right)}{\left( 1-\sin A \right)\left( 1+\sin A \right)}}$

Now, by applying the identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$, we get

\[\Rightarrow \sqrt{\dfrac{1+\sin A}{1-\sin A}}=\sqrt{\dfrac{{{\left( 1+\sin A \right)}^{2}}}{1-{{\sin }^{2}}A}}\]

Now, we know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, we get

\[\Rightarrow \sqrt{\dfrac{1+\sin A}{1-\sin A}}=\dfrac{1+\sin A}{\sqrt{{{\cos }^{2}}A}}\]

\[\Rightarrow \sqrt{\dfrac{1+\sin A}{1-\sin A}}=\dfrac{1+\sin A}{\cos A}\]

\[\Rightarrow \sqrt{\dfrac{1+\sin A}{1-\sin A}}=\dfrac{1}{\cos A}+\dfrac{\sin A}{\cos A}\]

\[\Rightarrow \sqrt{\dfrac{1+\sin A}{1-\sin A}}=\sec A+\tan A\]

\[\Rightarrow \sqrt{\dfrac{1+\sin A}{1-\sin A}}=RHS\]

$\therefore \sqrt{\dfrac{1+\sin A}{1-\sin A}}=\sec A+\tan A$ 

Hence proved

1. $\dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2\cos \theta -\cos \theta }=\tan \theta $ 

Ans: Given expression is $\dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2\cos \theta -\cos \theta }=\tan \theta $.

Let us consider the LHS of the given expression, we get

$LHS=\dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2\cos \theta -\cos \theta }$

Taking common terms out, we get

$\Rightarrow \dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2\cos \theta -\cos \theta }=\dfrac{\sin \theta \left( 1-2{{\sin }^{2}}\theta  \right)}{\cos \theta \left( 2{{\cos }^{2}}\theta -1 \right)}$

Now, we know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, we get

$\Rightarrow \dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2\cos \theta -\cos \theta }=\dfrac{\sin \theta \left( 1-2{{\sin }^{2}}\theta  \right)}{\cos \theta \left( 2\left( 1-2{{\sin }^{2}}\theta  \right)-1 \right)}$

$\Rightarrow \dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2\cos \theta -\cos \theta }=\dfrac{\sin \theta \left( 1-2{{\sin }^{2}}\theta  \right)}{\cos \theta \left( 2-2{{\sin }^{2}}\theta -1 \right)}$

$\Rightarrow \dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2\cos \theta -\cos \theta }=\dfrac{\sin \theta \left( 1-2{{\sin }^{2}}\theta  \right)}{\cos \theta \left( 1-2{{\sin }^{2}}\theta  \right)}$

$\Rightarrow \dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2\cos \theta -\cos \theta }=\dfrac{\sin \theta }{\cos \theta }$

$\Rightarrow \dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2\cos \theta -\cos \theta }=\tan \theta $

$\Rightarrow \dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2\cos \theta -\cos \theta }=RHS$

$\therefore \dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2\cos \theta -\cos \theta }=\tan \theta $

Hence proved

1. ${{\left( \sin A+cosecA \right)}^{2}}+{{\left( \cos A+secA \right)}^{2}}=7+{{\tan }^{2}}A+{{\cot }^{2}}A$ 

Ans: Given expression is ${{\left( \sin A+cosecA \right)}^{2}}+{{\left( \cos A+\sec A \right)}^{2}}=7+{{\tan }^{2}}A+{{\cot }^{2}}A$.

Let us consider the LHS of the given expression, we get

$LHS={{\left( \sin A+cosecA \right)}^{2}}+{{\left( \cos A+\sec A \right)}^{2}}$

Now, by applying the identity \[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\], we get

\[\Rightarrow {{\left( \sin A+cosecA \right)}^{2}}+{{\left( \cos A+\sec A \right)}^{2}}={{\sin }^{2}}A+cosec{{A}^{2}}+2\sin AcosecA+{{\cos }^{2}}A+{{\sec }^{2}}A+2\cos A\sec A\]\[\Rightarrow {{\left( \sin A+cosecA \right)}^{2}}+{{\left( \cos A+\sec A \right)}^{2}}={{\sin }^{2}}A+{{\cos }^{2}}A+cosec{{A}^{2}}+{{\sec }^{2}}A+2\sin AcosecA+2\cos A\sec A\]We know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, $cosec\theta =\dfrac{1}{\sin \theta }$ and $sec\theta =\dfrac{1}{\cos \theta }$, we get

\[\Rightarrow {{\left( \sin A+cosecA \right)}^{2}}+{{\left( \cos A+\sec A \right)}^{2}}=1+cose{{c}^{2}}\theta +{{\sec }^{2}}\theta +2\sin A\dfrac{1}{\sin A}+2\cos A\dfrac{1}{\cos A}\] 

\[\Rightarrow {{\left( \sin A+cosecA \right)}^{2}}+{{\left( \cos A+\sec A \right)}^{2}}=1+\left( 1+{{\cot }^{2}}A+1+{{\tan }^{2}}A \right)+2+2\]

\[\Rightarrow {{\left( \sin A+cosecA \right)}^{2}}+{{\left( \cos A+\sec A \right)}^{2}}=7+{{\tan }^{2}}A+{{\cot }^{2}}A\]

\[\Rightarrow {{\left( \sin A+cosecA \right)}^{2}}+{{\left( \cos A+\sec A \right)}^{2}}=RHS\]

\[\therefore {{\left( \sin A+cosecA \right)}^{2}}+{{\left( \cos A+\sec A \right)}^{2}}=7+{{\tan }^{2}}A+{{\cot }^{2}}A\]

Hence proved

1. $\left( cosecA-\sin A \right)\left( \sec A-\cos A \right)=\dfrac{1}{\tan A+\cot A}$

Ans: Given expression is $\left( cosecA-\sin A \right)\left( \sec A-\cos A \right)=\dfrac{1}{\tan A+\cot A}$.

Let us consider the LHS of the given expression, we get

$LHS=\left( cosecA-\sin A \right)\left( \sec A-\cos A \right)$

We know that $cosec\theta =\dfrac{1}{\sin \theta }$ and $sec\theta =\dfrac{1}{\cos \theta }$, we get

$\Rightarrow \left( cosecA-\sin A \right)\left( \sec A-\cos A \right)=\left( \dfrac{1}{\sin A}-\sin A \right)\left( \dfrac{1}{\cos A}-\cos A \right)$

$\Rightarrow \left( cosecA-\sin A \right)\left( \sec A-\cos A \right)=\left( \dfrac{1-{{\sin }^{2}}A}{\sin A} \right)\left( \dfrac{1-{{\cos }^{2}}A}{\cos A} \right)$

Now, we know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, we get

$\Rightarrow \left( cosecA-\sin A \right)\left( \sec A-\cos A \right)=\left( \dfrac{{{\cos }^{2}}A}{\sin A} \right)\left( \dfrac{{{\sin }^{2}}A}{\cos A} \right)$

$\Rightarrow \left( cosecA-\sin A \right)\left( \sec A-\cos A \right)=\sin A\cos A$

Now, consider the RHS of the given expression, we get

$RHS=\dfrac{1}{\tan A+\cot A}$

Now, we know that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and $\cot \theta =\dfrac{\cos \theta }{\sin \theta }=\dfrac{1}{\tan \theta }$.

$\Rightarrow \dfrac{1}{\tan A+\cot A}=\dfrac{1}{\dfrac{\sin A}{\cos A}+\dfrac{\cos A}{\sin A}}$

$\Rightarrow \dfrac{1}{\tan A+\cot A}=\dfrac{1}{\dfrac{{{\sin }^{2}}A+{{\cos }^{2}}A}{\sin A\cos A}}$

$\Rightarrow \dfrac{1}{\tan A+\cot A}=\dfrac{\sin A\cos A}{{{\sin }^{2}}A+{{\cos }^{2}}A}$

Now, we know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, we get

$\Rightarrow \dfrac{1}{\tan A+\cot A}=\sin A\cos A$

Here, we get LHS=RHS

$\therefore \left( cosecA-\sin A \right)\left( \sec A-\cos A \right)=\dfrac{1}{\tan A+\cot A}$ 

Hence proved

1. $\left( \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A} \right)={{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}$

Ans: Given expression is $\dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}={{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}$.

Let us consider the LHS of the given expression, we get

$LHS=\dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}$

By applying the identities ${{\sec }^{2}}A=1+{{\tan }^{2}}A$ and ${{\operatorname{cosec}}^{2}}A=1+{{\cot }^{2}}A$, we get

$\Rightarrow \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}=\dfrac{se{{c}^{2}}A}{{{\operatorname{cosec}}^{2}}A}$

We know that $cosec\theta =\dfrac{1}{\sin \theta }$ and $sec\theta =\dfrac{1}{\cos \theta }$, we get

$\Rightarrow \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}=\dfrac{\dfrac{1}{{{\cos }^{2}}A}}{\dfrac{1}{{{\sin }^{2}}A}}$

$\Rightarrow \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}=\dfrac{{{\sin }^{2}}A}{{{\cos }^{2}}A}$

$\Rightarrow \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}={{\tan }^{2}}A$

Now, consider the RHS of the given expression, we get

$RHS={{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}$

Now, we know that $\cot \theta =\dfrac{1}{\tan \theta }$, we get

$\Rightarrow {{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}={{\left( \dfrac{1-\tan A}{1-\dfrac{1}{\tan A}} \right)}^{2}}$

$\Rightarrow {{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}={{\left( \dfrac{1-\tan A}{\dfrac{\tan A-1}{\tan A}} \right)}^{2}}$

$\Rightarrow {{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}={{\left( -\tan A \right)}^{2}}$

$\Rightarrow {{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}={{\tan }^{2}}A$

Here, we get LHS=RHS

$\therefore \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}={{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}$ 

Hence proved