Question

I need help with this practice problem solving I will send two additional pictures that go with this, one is the rest of the question and the other is the answer options

Answer 1

Given:

$\frac{\sec x\sin x}{\tan x+\cot x}=\sin^2x$

Find-: Prove the trigonometric identity.

Sol:

Use some trigonometric formula:

$\begin{gathered} \sec x=\frac{1}{\cos x} \\ \\ \tan x=\frac{\sin x}{\cos x} \\ \\ \cot x=\frac{\cos x}{\sin x} \end{gathered}$

So,

$\begin{gathered} \frac{\sec x\sin x}{\tan x+\cot x}=\sin^2x \\ \\ \frac{\frac{1}{\cos x}\cdot\sin x}{\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}}.................\text{ First option } \end{gathered}$

Then,

$=\frac{\frac{\sin x}{\cos x}}{\frac{\sin^2x}{\sin x\cos x}+\frac{\cos^2x}{\sin x\cos}}................................(\text{ Second option\rparen}$

$=\frac{\frac{\sin x}{\cos x}}{\frac{\sin^2x+\cos^2x}{\sin x\cos x}}...........................(\text{ Third option\rparen}$

Solve the identity then,

$=\frac{\frac{\sin x}{\cos x}}{\frac{1}{\sin x\cos x}}..................(\text{ Fourth option\rparen}$

Here, use the formula:

$\sin^2x+\cos^2x=1$

Then,

$\begin{gathered} =\frac{\sin x}{\cos x}\sin x\cos x..........(\text{ Fifth option\rparen} \\ \\ =\sin^2x \end{gathered}$