Question

Evaluate: sin(90-theta)cosec(90-theta)cot theta/sec(90-theta)cos(90-theta)tan(90-theta)+sin^265°+sin^225/tan10tan35tan55tan80+2cos^232cosec^258-cos^45

Answer 1

Answer:

sin theta ×cosec theta × tan theta ÷sec theta cos theta ×cot theta

=tan theta /cot theta

=tan² theta..

Answer 2

Given:

$\bold{\frac{\sin(90-\theta)\ cosec(90-\theta)\cot \theta}{\sec(90-\theta)\cos(90-\theta)tan(90-\theta)}+\frac{\sin^2{65} +\sin^2{25}}{\tan10\tan35\tan55\tan80}+2\cos^{2}32\ cosec^{2}58\ - \cos^4{5}}$To prove:

Evaluate.

Solution:

Formula:

$\bold{\sin^2 \theta+ \cos^2 \theta = 1 }\\\\\bold{\tan10\tan35\tan55\tan80 = 1} \\\\ \bold{\cos 5 = 0.28366 \ \ So, \cos^4 5= 0.00647\dots }$

$\Rightarrow \frac{\sin(90-\theta)\ cosec(90-\theta)\cot \theta}{\sec(90-\theta)\cos(90-\theta)tan(90-\theta)}+\frac{\sin^2{65} +\sin^2{25}}{\tan10\tan35\tan55\tan80}+2\cos^{2}32\ cosec^{2}58\ - \cos^4{5}$

$= \frac{\cos \theta\sec\theta\cot \theta}{\ cosec \theta\sin\theta \cot\theta}+\frac{\sin^2{65} +\sin^2{25}}{\tan10\tan35\tan55\tan80}+2\cos^{2}32\ cosec^{2}58\ - \cos^4{5}}$

$= 1 +\frac{\sin^2{65} +\sin^2{25}}{\tan10\tan35\tan55\tan80}+2\cos^{2}32\ cosec^{2}58\ - \cos^4{5}}$

$= 1 +\frac{\sin^2{65} +\sin^2{(90- 65)}}{\tan10\tan35\tan55\tan80}+2\cos^{2}(90-32)\ cosec^{2}58\ - \cos^4{5}}$

$= 1 +\frac{\sin^2{65} +\cos^2{25}}{\tan10\tan35\tan55\tan80}+2\sin^{2}58\ \frac{1}{\sin^{2}58}\ - \cos^4{5}}$

$= 1 +\frac{1}{1}+2\times 1 \ - 0.00647}$

$= 1 +1+2 \ - 0.00647\\\\=4+0.00647\\\\=4.00647 \ \ \ \ or \ = 4$

The evaluated value is = 4.