Question

answer me fast if you know...if you don't know so don't do anything.​

Answer 1

Step-by-step explanation:

$\sf{LHS=\frac{1}{\sec(x)-\tan(x)}-\frac{1}{\cos(x)}}$

$\sf{=\frac{1}{\frac{1}{\cos(x)}-\frac{\sin(x)}{\cos(x)}}-\frac{1}{\cos(x)}}$

$\sf{=\frac{1}{\frac{1-\sin(x)}{\cos(x)}}-\frac{1}{\cos(x)}}$

$\sf{=\frac{\cos(x)}{1-\sin(x)}-\frac{1}{\cos(x)}}$

$\sf{=\frac{\cos^{2}(x)-[1-\sin(x)]}{[1-\sin(x)]\cos(x)}}$

$\sf{=\frac{[1-\sin^{2}(x)]-[1-\sin(x)]}{[1-\sin(x)]\cos(x)}}$

$\sf{=\frac{[(1-\sin(x))(1+\sin(x))]-[1-\sin(x)]}{[1-\sin(x)]\cos(x)}}$

$\sf{=\frac{[(1-\sin(x))][(1+\sin(x)-1)]}{[1-\sin(x)]\cos(x)}}$

$\sf{=\frac{(1+\sin(x)-1)}{\cos(x)}}$

$\sf{=\frac{\sin(x)}{\cos(x)}}$

$\sf{=\tan(x) ......(i)}$

$\sf{RHS=\frac{1}{\cos(x)}-\frac{1}{\sec(x)+\tan(x)}}$

$\sf{=\frac{1}{\cos(x)}-\frac{1}{\frac{1}{\cos(x)}+\frac{\sin(x)}{\cos(x)}}}$

$\sf{=\frac{1}{\cos(x)}-\frac{\cos(x)}{1+\sin(x)}}$

$\sf{=\frac{[1+\sin(x)]-\cos^{2}(x)}{\cos(x)[1+\sin(x)]}}$

$\sf{=\frac{[1+\sin(x)]-[1-\sin^{2}(x)]}{\cos(x)[1+\sin(x)]}}$

$\sf{=\frac{[1+\sin(x)]-[(1-\sin(x))(1+\sin(x))]}{\cos(x)[1+\sin(x)]}}$

$\sf{=\frac{[1+\sin(x)][1-1+\sin(x)]}{\cos(x)[1+\sin(x)]}}$

$\sf{=\frac{\sin(x)}{\cos(x)}}$

$\sf{=\tan(x) ......(ii)}$

From (i) & (ii), LHS = RHS... Hence proved

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