Question

secant x minus tangent X ka whole square is equals to 1 minus sin x upon 1 + sin x​

Answer 1

Step-by-step explanation:

LHS = ( sec x - tan x )^2

( 1/cos x - sin x/cos x )^2

[(1 - sin x)/cos x ]^2

(1 - sin x)^2 / cos^2 x

(1 - sin x)^2 / 1 - sin^2 x

(1 - sin x)^2 / (1 + sin x)(1 - sin x)

(1 - sin x ) / ( 1 + sin x) = RHS

Answer 2

$(\sec x-\tan x)^2=\dfrac{1-\sin x}{1+\sin x}$ Proved.

Step-by-step explanation:

Consider the provided information.

$(\sec x-\tan x)^2=\dfrac{1-\sin x}{1+\sin x}$

Consider the RHS.

$=\dfrac{1-\sin x}{1+\sin x}\times\dfrac{1-\sin x}{1-\sin x}$

$=\dfrac{(1-\sin x)^2}{1-\sin^2 x}$

$=\dfrac{(1-\sin x)^2}{\cos^2 x}$

$=\left(\dfrac{1-\sin x}{\cos x}\right)^2$

$=\left(\dfrac{1}{\cos x}-\dfrac{\sin x}{\cos x}\right)^2$

$=\left(\sec x}-\tan x\right)^2$

LHS=RHS

Hence, proved.

#Learn more

Prove that 1 + cos theta upon 1 minus cos theta is equal to cosec theta + cot theta ka whole square​

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