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The value of (sec A + tan A) (1 - sin A) is:
Question

The value of (sec A + tan A) (1 - sin A) is:

A.

​tan⁡ A

B.

cot⁡ A

C.

​ sin⁡ A

D.

cos ⁡A

Correct option is D

Given:

Expression = (secA+tanA)(1sinA)(\sec A + \tan A)(1 - \sin A)​​

Formula Used:

secA=1cosA\sec A = \frac{1}{\cos A}​​

tanA=sinAcosA\tan A = \frac{\sin A}{\cos A}​​

Solution:

(secA+tanA)(1sinA) =(1+sinAcosA)(1sinA)(\sec A + \tan A)(1 - \sin A) \\ \ \\= \left( \frac{1 + \sin A}{\cos A} \right)(1 - \sin A)​​

Now apply identity:

(1+sinA)(1sinA)=1sin2A=cos2A(1 + \sin A)(1 - \sin A) = 1 - \sin^2 A = \cos^2 A​​

So,

1+sinAcosA×(1sinA) =cos2AcosA=cosA\frac{1 + \sin A}{\cos A} \times (1 - \sin A) \\ \ \\ = \frac{\cos^2 A}{\cos A} = \cos A​​

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