Question

(secA+tanA)(1-sinA) please send the answer please fast​

Answer 1

Answer:

cos A

Step-by-step explanation:

(1--sin^2A)/cosA=cosA

Answer 2

ANSWER:

To Solve:

• (secA + tanA)(1 - sinA)

Solution:

We are given that,

$\implies(\sec A+\tan A)(1-\sin A)$

We know that,

$\hookrightarrow\sec\theta=\dfrac{1}{\cos\theta}$

So,

$\implies(\sec A+\tan A)(1-\sin A)$

$\implies\bigg(\dfrac{1}{\cos A}+\tan A\bigg)\bigg(1-\sin A\bigg)$

We also know that,

$\hookrightarrow\tan\theta=\dfrac{\sin\theta}{\cos\theta}$

So,

$\implies\bigg(\dfrac{1}{\cos A}+\tan A\bigg)\bigg(1-\sin A\bigg)$

$\implies\bigg(\dfrac{1}{\cos A}+\dfrac{\sin A}{\cos A}\bigg)\bigg(1-\sin A\bigg)$

Taking LCM,

$\implies\bigg(\dfrac{1+\sin A}{\cos A}\bigg)\bigg(1-\sin A\bigg)$

On simplifying,

$\implies\bigg(\dfrac{(1+\sin A)(1-\sin A)}{\cos A}\bigg)$

We know that,

$\hookrightarrow(x-y)(x+y)=x^2-y^2$

Hence,

$\implies\bigg(\dfrac{(1+\sin A)(1-\sin A)}{\cos A}\bigg)$

$\implies\bigg(\dfrac{1^2-\sin^2A)}{\cos A}\bigg)$

$\implies\bigg(\dfrac{1-\sin^2A}{\cos A}\bigg)$

We know that,

$\hookrightarrow1-\sin^2\theta=\cos^2\theta$

So,

$\implies\bigg(\dfrac{1-\sin^2A}{\cos A}\bigg)$

$\implies\bigg(\dfrac{\cos^2A}{\cos A}\bigg)$

$\implies\dfrac{\cos^2A}{\cos A}$

Dividing in numerator and denominator by cos A,

$\bf\implies cos A$