Trigonometric Ratios And Identities - Trigonometry - Mathematics Previous Year Questions

\({\sin ^2}17.5^\circ + \sin 72.5^\circ\) is equal to

If x and y are acute angles, such that \(\cos x + \cos y = {3 \over 2}\) and \(\sin x + \sin y = {3 \over 4}\), then \(\sin (x + y)\) equals

The expression \({{\tan A} \over {1 - \cot A}} + {{\cot A} \over {1 - \tan A}}\) can be written as

If \(\sin A+\sin B=a\) and \(\cos A+\cos B=b\), then \(\cos (A+B)\) equals?

What is \({{\cos \theta } \over {1 - \tan \theta }} + {{\sin \theta } \over {1 - \cot \theta }}\) equal to?

$$\cos ^6 A-\sin ^6 A \text { is equal to }$$

$$\text { If } \operatorname{cosec}(90+A)+x \cos A \cot (90+A)=\sin (90+A) \text { then the value of } x \text { is }$$

If \(\cos \alpha=k \cos \beta\) then \(\cot \left(\frac{\alpha+\beta}{2}\right)\) is equal to

$$\text { If } \frac{x}{\cos \theta}=\frac{y}{\cos \left(\theta+\frac{2 \pi}{3}\right)}=\frac{z}{\cos \left(\theta-\frac{2 \pi}{3}\right)} \text { then } x+y+z \text { is equal to }$$

$$\left(\cos \frac{\pi}{12}-\sin \frac{\pi}{12}\right)\left(\tan \frac{\pi}{12}+\cot \frac{\pi}{12}\right)=$$

$$\frac{\cos 9^{\circ}+\sin 9^{\circ}}{\cos 9^{\circ}-\sin 9^{\circ}}=$$

$$\text { Value of } \cos 105^{\circ} \text { is }$$

$$\sqrt{2+\sqrt{2+\sqrt{2+2 \cos 8 \theta}}} \text { where } \theta \in\left[-\frac{\pi}{8}, \frac{\pi}{8}\right] \text { is equal to }$$

$$\text { If } \frac{\cos x}{\cos (x-2 y)}=\lambda \text { then } \tan (x-y) \tan y=$$

If \(\cos \theta=\frac{1}{2}\left(x+\frac{1}{x}\right)\) then \(\frac{1}{2}\left(x^2+\frac{1}{x^2}\right)=\)

$$4\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{5 \pi}{8}\right)\left(1+\cos \frac{7 \pi}{8}\right) \text { is equal to }$$

$$\text { If } \sin A=\frac{4}{5} \text { and } \cos B=\frac{-12}{13} \text { where } A \text { and } B \text { lie in first and third quadrant respectively. Then } \cos (A+B)=$$

Simplified expression of

$1-\frac{\sin ^2 y}{1+\cos y}+\frac{1+\cos y}{\sin y}-\frac{\sin y}{1-\cos y}$ is :