Question

What is $\frac{{\cos }^2({45}^o+\theta )+{\cos }^2({45}^o-\theta )}{\tan ({60}^o+\theta )+\tan ({30}^o-\theta )}$ equal to?

• A. $-1$
• B. $0$
• C. $1$
• D. $2$

Answer 1

The correct option is C $1$

Let us consider the problem:

$\frac{{\cos }^2\left({45}^0+\theta \right)+{\cos }^2\left({45}^0-\theta \right)}{\tan \left({60}^0+\theta \right)\tan \left({30}^0-\theta \right)}$

$=\frac{{\cos }^2\left(45+\theta \right)+{\sin }^2\left({90}^0-{45}^0+\theta \right)}{\tan \left({60}^0+\theta \right)\cot \left({90}^0-{30}^0+\theta \right)}$

$=\frac{{\cos }^2\left({45}^0+\theta \right)+{\sin }^2\left({45}^0+\theta \right)}{\tan \left({60}^0+\theta \right)\times \cot \left({60}^0+\theta \right)}$

$=\frac{1}{1}$ $\left(\therefore {\sin }^2\theta +{\cos }^2\theta =1,\tan \theta cot\theta =1\right)$

$=1$