Question

prove (sec A + tan A)(1-sin A) =cos A​

Answer 1

Prove that (sec A + tan A)(1 – sin A) = cos A. (a + b)(a – b)= a2 – b2, So, (1 + sin A)(1 – sin A) = 1 – sin2A. Hence, the value of (sec A + tan A) (1 – sin A) is equal to cos A.

Answer 2

Step-by-step explanation:

LHS = (secA + tanA)(1 – sinA)

RHS = cosA

Solving LHS,

$\rm{\longrightarrow} \: (secA + tanA)(1 - sinA)$

$\rm{\longrightarrow} \: \bigg( \dfrac{1}{cosA} + \dfrac{sinA}{cosA}\bigg)(1 - sinA)$

$\rm{\longrightarrow} \: \bigg( \dfrac{1 + sinA}{cosA}\bigg)(1 - sinA)$

$\rm{\longrightarrow} \: \dfrac{1 + sinA - {sinA}^{2} -sinA }{cosA}$

$\rm{\longrightarrow} \: \dfrac{1 - {sinA}^{2} }{cosA}$

Using the trignometric identity,

sinA² + cosA² = 1

$\rm{\longrightarrow} \: \dfrac{{cosA}^{2}}{cosA}$

$\rm{\longrightarrow} \: cosA$

LHS = cosA

RHS = cosA

LHS = RHS

Hence proved.