Question

# The expression $$\sin A . \cos A \left (\dfrac{\tan^{2}\ A}{\tan A - 1} + \dfrac{\cot^2 A}{\cot A - 1} \right )$$ can be written as:

A
cosAcosecA+1
B
tanA+secA
C
secA+cosecA
D
sinAcosA+1

Solution

## The correct option is D $$\sin A \cos A + 1$$ $$\dfrac{\tan^{2}\ A}{\tan A - 1} + \dfrac{\cot^2 A}{\cot A - 1}$$$$= \dfrac{\tan A}{\dfrac{(\tan A - 1)}{\tan A}} + \dfrac{\cot A}{\dfrac{(\cot A - 1)}{\cot A}}$$$$= \dfrac{\tan A}{1 - \cot A} + \dfrac{\cot A}{1 - \tan A}$$$$= \dfrac{\tan^2 A}{\tan A - 1} - \dfrac{\cot A}{\tan A - 1}$$$$= \dfrac{\tan^2 A - \cot A}{\tan A - 1}$$ $$= \dfrac{\tan^3 A - 1}{\tan^2 A - \tan A}$$$$= \dfrac{(\tan A - 1) (\tan^2 A + \tan A + 1)}{\tan A (\tan A - 1)}$$ $$= \dfrac{\sec^2 A + \tan A}{\tan A }$$ $$= \dfrac{1}{\sin A \cos A} + 1$$  $$\therefore \sin A \cdot \cos A \left (\dfrac{\tan^{2}\ A}{\tan A - 1} + \dfrac{\cot^2 A}{\cot A - 1} \right ) = 1 + \sin A \cos A$$Mathematics

Suggest Corrections

0

Similar questions
View More

People also searched for
View More