Question

$\left\{\frac{\cos 65°}{\sin 25°}+\frac{\mathrm{cosec}34°}{\sec 56°}-\frac{2\cos 43°\mathrm{cosec}47°}{\tan 10°\tan 40°\tan 50°\tan 80°}\right\}$

Answer 1

$$\begin{gathered} \left\{\frac{\cos 65°}{\sin 25°}+\frac{\mathrm{cosec}34°}{\sec 56°}-\frac{2\cos 43°\mathrm{cosec}47°}{\tan 10°\tan 40°\tan 50°\tan 80°}\right\} \\ =\left\{\frac{\cos \left(90°-25°\right)}{\sin 25°}+\frac{\mathrm{cosec}34°}{\sec 56°}-\frac{2\cos \left(90°-47°\right)\mathrm{cosec}47°}{\tan 10°\tan 40°\tan 50°\tan 80°}\right\} \\ =\left\{\frac{\sin 25°}{\sin 25°}+\frac{\mathrm{cosec}34°}{\sec 56°}-\frac{2\sin 47°\mathrm{cosec}47°}{\tan 10°\tan 40°\tan 50°\tan 80°}\right\}\left(\because \cos \left(90°-\theta \right)=\sin \theta \right) \\ =\left\{1+\frac{\mathrm{cosec}\left(90°-56°\right)}{\sec 56°}-\frac{2\sin 47°\mathrm{cosec}47°}{\tan 10°\tan 40°\tan 50°\tan 80°}\right\} \\ =\left\{1+\frac{\sec 56°}{\sec 56°}-\frac{2\sin 47°\mathrm{cosec}47°}{\tan 10°\tan 40°\tan 50°\tan 80°}\right\}\left(\because \mathrm{cosec}\left(90°-\theta \right)=\sec \theta \right) \\ =\left\{1+1-\frac{2\sin 47°{\displaystyle \frac{1}{\sin 47°}}}{\tan 10°\tan 40°\tan 50°\tan 80°}\right\}\left(\because \mathrm{cosec}\theta =\frac{1}{\sin \theta }\right) \\ =\left\{2-\frac{2}{\tan 10°\tan 40°\tan 50°\tan 80°}\right\} \\ =\left\{2-\frac{2}{\tan 10°\tan 40°\tan \left(90°-40°\right)\tan \left(90°-10°\right)}\right\} \\ =\left\{2-\frac{2}{\tan 10°\tan 40°\cot 40°\cot 10°}\right\}\left(\because \tan \left(90°-\theta \right)=\cot \theta \right) \\ =\left\{2-\frac{2}{\tan 10°\tan 40°{\displaystyle \frac{1}{\tan 40°}}{\displaystyle \frac{1}{\tan 10°}}}\right\}\left(\because \cot \theta =\frac{1}{\tan \theta }\right) \\ =\left\{2-\frac{2}{1}\right\} \\ =2-2 \\ =0 \\ \mathrm{Hence},\left\{\frac{\cos 65°}{\sin 25°}+\frac{\mathrm{cosec}34°}{\sec 56°}-\frac{2\cos 43°\mathrm{cosec}47°}{\tan 10°\tan 40°\tan 50°\tan 80°}\right\}=0. \end{gathered}$$