Question

(sin²31° + sin²59° / sec²35° - cot²55° + tan29°cot60° - cosec²61°)

Answer 1

Answer:

please mark as brainloest

Answer 2

The answer is $\frac{1}{2}$

Step-by-step explanation:

Given,

$\frac{sin^{2}31+sin^{2}59}{sec^{2}35-cot^{2}55+tan29.cot61-cosec^{2}61}=?$

∴ $\frac{sin^{2}31+sin^{2}59}{sec^{2}35-cot^{2}55+tan29.cot61-cosec^{2}61}$

$=\frac{sin^{2}31+sin^{2}(90-31)}{sec^{2}35-cot^{2}(90-35)+tan(90-61).cot61-cosec^{2}61}$

$=\frac{(sin^{2}31+cos^{2}31)}{(sec^{2}35-tan^{2}35)+(cot^{2} 61-cosec^{2}61)}$  where $sin^{2}(90-\theta)=cos^{2}\theta,cot^{2}(90-\theta)=tan^{2}\theta\ and\ tan(90-\theta)=cot\theta$

$=\frac{1}{1+1}$

$=\frac{1}{2}$

So, The answer is $\frac{1}{2}$