Question

The value of $\tan \theta .\tan ({90}^o-\theta )+\cos \theta .\text{cosec}({90}^o-\theta )$ is

• A. $0$
• B. $-1$
• C. $1$
• D. $2$

Answer 1

Answer: (D)

$2$

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

and $\text{cosec}(90-\theta )=\sec \theta$

Also $\tan \theta =\frac{1}{\cot \theta }$,

$\cos \theta =\frac{1}{\sec \theta }$

So, $\tan \theta .\tan ({90}^o-\theta )+\cos \theta .\text{cosec}({90}^o-\theta )$

$=\tan \theta \cot \theta +\cos \theta \sec \theta$

$=1+1=2$