Question

Find the value of $\cos ({90}^o-A)\tan ({90}^o-A)\sec ({90}^o-A)$

• A. $\cot A$
• B. $\tan A$
• C. $\cos A$
• D. $\text{cosec}A$

Answer 1

The correct option is A $\cot A$

$\cos ({90}^o-A).\tan ({90}^o-A)\sec ({90}^o-A)$

$=\sin A\times \cot A\times \text{cosec}A$ ..... $[\because \cos ({90}^o-\theta )=\sin \theta ,\tan ({90}^o-\theta )=\cot \theta ,\sec ({90}^o-\theta )=\text{cosec}\theta ]$

$=\sin A\times \frac{\cos A}{\sin A}\times \frac{1}{\sin A}$

$=\frac{\cos A}{\sin A}=\cot A$

Hence, option A is correct.