Question

Simplify : $\frac{{\text{cosec}}^2{67}^o-{\tan }^2{23}^o}{{\sec }^2{20}^o-{\cot }^2{70}^o}$

Answer 1

We have,$\frac{cose{c}^{2}67-ta{n}^{2}23}{se{c}^{2}20-co{t}^{2}70}=\frac{1+co{t}^{2}67-ta{n}^{2}23}{1+ta{n}^{2}20-co{t}^{2}70}$

We know that,

$1+co{t}^2\theta =cose{c}^2\theta$

$1+ta{n}^2\theta =se{c}^2\theta$

Now, $tan(90-\theta )=cot\theta$

$cot(90-\theta )=tan\theta$

$=\frac{1+co{t}^2(90-23)-ta{n}^{2}23}{1+ta{n}^2(90-70)-co{t}^{2}70}$

$=\frac{1+ta{n}^{2}23-ta{n}^{2}23}{1+co{t}^{2}70-co{t}^{2}70}$

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