Question

(sec A + tan A)(1-sin A) = cos A​

Answer 1

Step-by-step explanation:

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

=>secA-secA×sinA+tanA-tanA×sinA=cosA

=>secA-tanA+tanA-sin²A/cosA=cosA

=>secA-sin²A/cosA=cosA

=>1-sin²A/cosA=cosA

=>cos²A/cosA=cosA

=>cosA=cosA

Answer 2

Solution!!

$\sf (\sec A +\tan A)(1-\sin A)=\cos A$

Taking LHS,

$\sf =(\sec A +\tan A)(1-\sin A)$

Use $\sf \sec \theta =\dfrac{1}{\cos \theta }$ to transform the expression.

$\sf =\left(\dfrac{1}{\cos A}+\tan A\right)(1-\sin A)$

Use $\sf \tan \theta =\dfrac{\sin \theta }{\cos \theta }$ to transform the expression.

$\sf =\left(\dfrac{1}{\cos A}+\dfrac{\sin A}{\cos A}\right)(1-\sin A)$

Take the LCM.

$\sf =\left(\dfrac{1+\sin A}{\cos A}\right)(1-\sin A)$

Calculate the product.

$\sf =\dfrac{(1+\sin A)(1-\sin A)}{\cos A}$

Use (a + b)(a - b) = a² - b² to simplify the expression.

$\sf =\dfrac{1-\sin ^{2}A}{\cos A}$

Use $\sf 1-\sin ^{2}\theta =\cos ^{2}\theta$ to simplify the expression.

$\sf =\dfrac{\cos ^{2}A}{\cos A}$

Simplify the expression.

$\sf =\cos A$

LHS = RHS

Hence, proved.