Chapter 8: Introduction to Trigonometry & Its Equations

According to the question,

cosec θ + cot θ = p

Since,

Hence, proved.

L.H.S=

√ (sec² θ + cosec² θ)

Since,

= R.H.S

Hence, proved.

Let PR = h meter, be the height of the tower.

The observer is standing at point Q such that, the distance between the observer and tower is QR = (20+x) m, where

QR = QS + SR = 20 + x

∠PQR = 30°

∠ PSR = θ

In ∆PQR,

Rearranging the terms,

We get 20 +x = √3h

⇒ x = √3h – 20 …eq.1

In ∆PSR,

tan θ = h/x

Since, angle of elevation increases by 15^o when the observer moves 20 m towards the tower.

We have,

θ = 30° + 15° = 45°

So,

tan 45^o = h/x

⇒ 1 = h/x

⇒ h = x

Substituting x=h in eq. 1, we get

h = √3 h – 20

⇒ √3 h – h = 20

⇒ h (√3 – 1) = 20

= 10 (√3 + 1)

Hence, the required height of the tower is 10 (√3 + 1) meter.

Given: 1+sin² θ = 3 sin θ cos θ

Dividing L.H.S and R.H.S equations with sin² θ,

We get,

cosec² θ + 1 = 3 cot θ

Since,

cosec² θ – cot² θ = 1 ⇒ cosec² θ = cot² θ +1

⇒ cot² θ +1+1 = 3 cot θ

⇒ cot² θ +2 = 3 cot θ

⇒ cot² θ –3 cot θ +2 = 0

Splitting the middle term and then solving the equation,

⇒ cot² θ – cot θ –2 cot θ +2 = 0

⇒ cot θ(cot θ -1)–2(cot θ +1) = 0

⇒ (cot θ – 1)(cot θ – 2) = 0

⇒ cot θ = 1, 2

Since,

tan θ = 1/cot θ

tan θ = 1, ½

Hence, proved.

Given: sin θ +2 cos θ = 1

Squaring on both sides,

(sin θ +2 cos θ)² = 1

⇒ sin² θ + 4 cos² θ + 4sin θcos θ = 1

Since, sin² θ = 1 – cos² θ and cos² θ = 1 – sin² θ

⇒ (1 – cos² θ) + 4(1 – sin² θ) + 4sin θcos θ = 1

⇒ 1 – cos² θ + 4 – 4 sin² θ + 4sin θcos θ = 1

⇒ – 4 sin² θ – cos² θ + 4sin θcos θ = – 4

⇒ 4 sin² θ + cos² θ – 4sin θcos θ = 4

We know that,

a²+ b² – 2ab = ( a – b)²

So, we get,

(2sin θ – cos θ)² = 4

⇒ 2sin θ – cos θ = 2

Hence proved.

Let BC = s; PC = t

Let height of the tower be AB = h.

∠ABC = θ and ∠APC = 90° – θ

(∵ the angle of elevation of the top of the tower from two points P and B are complementary)

⇒ h² = st

⇒ h = √st

Hence the height of the tower is √st.

Let SQ = h be the tower.

∠SPQ = 30° and ∠SRQ = 60°

According to the question, the length of shadow is 50 m long hen angle of elevation of the sun is 30° than when it was 60°. So,

PR = 50 m and RQ = x m

So in ∆SRQ, we have

⇒ 50√3+h = 3h

⇒ 50√3 = 3h – h

⇒ 3h – h = 50√3

⇒ 2h = 50√3

⇒ h = (50√3)/2

⇒ h = 25√3

Hence, the required height is 25√3 m.

Given that a vertical flag staff of height h is surmounted on a vertical tower of height H(say),

such that FP = h and FO = H.

The angle of elevation of the bottom and top of the flag staff on the plane is ∠PRO = α and ∠FRO = β respectively

In ∆PRO, we have

Hence, proved.

Given: tan θ+ sec θ = l …eq. 1

Multiplying and dividing by (sec θ – tan θ) on numerator and denominator of L.H.S,

So, sec θ – tan θ = 1 …eq.2

Adding eq. 1and eq. 2, we get

(tan θ + sec θ) + (sec θ – tan θ) = 1

Hence, proved.

Class 10 students are advised to solve sample papers and previous years’ question papers to have an idea of the types of questions asked from the chapter Trigonometry. Also, find our online reading materials, such as notes, exemplar books, 10th Class Maths NCERT Solutions and question papers in downloadable PDF format to have a quick revision before the board exam.

Download BYJU’S – The Learning App to get personalised video lessons explaining Maths topics. Experience a new method of learning at BYJU’S to understand concepts effortlessly.