Trigonometric Ratios And Identities - Trigonometry - Mathematics Previous Year Questions

$$\frac{e^{4 x}+e^{-4 x}+14}{4\left(e^x-e^{-x}\right)^2}=$$

If $\tanh x=\frac{1}{2}$, then $\sinh 2 x-\operatorname{sech} 2 x=$

If $\cos x+\cos y=p, \sin x+\sin y=q$, then $\cos \left(\frac{x-y}{2}\right)=$

If $A+B+C=\frac{3 \pi}{2}$, then $4 \sin A \sin B \sin C+\cos 2 A+\cos 2 B+\cos 2 C=$

If $A$ and $B(A>B)$ are acute angles, $\sin (A-B)=\frac{16}{65}$ and $\sin B=\frac{5}{13}$, then $\tan A+\cot A=$

If $\tan A=\frac{2}{3}$, then $\sin 4 A=$

$$\frac{\sqrt{2} \cos 45^{\circ}+\cos 56^{\circ}+\cos 58^{\circ}-\cos 66^{\circ}}{\sqrt{2} \cos 28^{\circ} \cos 29^{\circ} \sin 33^{\circ}}$$

If $\theta=\frac{\pi}{12}$ and $x=\log \left(\cot \left(\frac{\pi}{4}+\theta\right)\right)$, then $\cosh x=$

$2 \cosh (x+y) \sinh (x-y)+\sinh 2 y=$

$$\text { If } \frac{2 \sin \theta}{1+\cos \theta+\sin \theta}=y, \text { then } \frac{1-\cos \theta+\sin \theta}{1+\sin \theta}=$$

If $\cos \frac{\pi}{7} \cos \frac{2 \pi}{7} \cos \frac{4 \pi}{7}=\frac{\sin \frac{8 \pi}{7}}{8 \sin \frac{\pi}{7}}$, then $\sin \frac{\pi}{14} \sin \frac{3 \pi}{14} \sin \frac{5 \pi}{14} \sin \frac{7 \pi}{14} \sin \frac{9 \pi}{14} \sin \frac{11 \pi}{14} \sin \frac{13 \pi}{14}=$

If $f(\theta)=\cos ^3 \theta+\cos ^3\left(\frac{2 \pi}{3}+\theta\right)+\cos ^3\left(\theta-\frac{2 \pi}{3}\right)$, then $f\left(\frac{\pi}{5}\right)=$

$$\sin 6^{\circ}+\sin 54^{\circ}+\sin 126^{\circ}+\cos 156^{\circ}=$$

If $\tan \alpha=\frac{-12}{5}, \cot \beta=\frac{7}{24}, \alpha$ does not belong to second quadrant and $\beta$ does not belong to first quadrant, then $\sqrt{13} \sin \frac{\alpha}{2}+\cos \frac{\beta}{2}+\tan \frac{\alpha}{2} \cot \frac{\beta}{2}=$

$\cos \frac{\pi}{7} \cos \frac{2 \pi}{7} \cos \frac{3 \pi}{7} \cos \frac{\pi}{14} \cos \frac{3 \pi}{14} \cos \frac{5 \pi}{14}=$

If $A+B+C+D=2 \pi$, then $\cos A-\cos B+\cos C-\cos D=$

If $540^{\circ}<\theta<630^{\circ}$ and $\tan \theta=5 / 12$, then

$$\frac{\cos \frac{\theta}{2}-5 \sin \frac{\theta}{2}}{\sqrt{-(12 \sec \theta+5 \operatorname{cosec} \theta)}}=$$

If $\cot \theta=-\frac{2}{3}$ and $\theta$ does not lie in the 4 th quadrant, then $\frac{(5 \sin \theta+\cos \theta)^2}{\tan \theta+\cot \theta}=$

If $\cosh x=\frac{4}{3}$, then $3 \cosh x+3^2 \cosh 2 x+3^3 \cosh 3 x=$

If $\frac{\sin ^4 x}{2}+\frac{\cos ^4 x}{3}=\frac{1}{5}$, then $27 \sec ^6 \alpha+8 \operatorname{cosec}^6 \alpha=$