Question

(sec A + tan A) (1- sin A) =​

Answer 1

Answer:

answer for the given problem is given

Answer 2

$\huge\sf\pink{Answer}$

☞ (sec A + tan A)(1-sin A) = cos A

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$\huge\sf\blue{Given}$

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

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$\huge\sf\gray{To \:Find}$

◈ The simplified form of the given Equation?

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$\huge\sf\purple{Steps}$

☯ $\underline{\textsf {As Per the Question}}$

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

◕ $\sf sec \ A = \dfrac{1}{cos \ A}$

◕ $\sf tan \ A = \dfrac{1}{sin \ A}{cos \ A}$

Substituting these values,

➝ $\sf \bigg\lgroup\dfrac{1}{cos \ A} + \dfrac{sec \ A}{cos \ A}\bigg\rgroup(1-sin \ A)$

➝ $\sf \bigg\lgroup\dfrac{1+sin \ A}{cos \ A}\bigg\rgroup (1-sin \ A)$

➝ $\sf \bigg\lgroup \dfrac{(1+sin \ A)(1-sin \ A)}{cos \ A}\bigg\rgroup$

➝ $\sf \dfrac{1^2-sin^2 \ A}{cos \ A}$ «« (a+b)(a-b) = a²-b² »»

➝ $\sf \dfrac{1-sin^2 \ A}{cos \ A}$

➝ $\sf \dfrac{cos^2 \ A}{cos \ A}$ «« 1-sin² A = cos² A »»

➝ $\sf\orange{(sec \ A + tan \ A)(1-sin \ A) = cos \ A}$

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