Question

prove that:

Cos A by 1 + sin a + 1 + sin a by Cos A the whole into Cos A by 1 - sin A - 1 - sin by Cos A equal to 4 tan A SecA.
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Answer 1

Answer:

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Answer 2

Answer with Step -by-step explanation:

LHS

$\frac{cosA}{1+sinA}+\frac{1+sinA}{cos A}(\frac{cosA}{1-sinA}-\frac{1-sinA}{cosA})$

$\frac{cos^2A(1+sinA)^2}{cosA(1+sinA)}\times (\frac{cos^2A-(1-sinA)^2}{cosA(1-sinA)})$

$\frac{cos^2A+1+sin^2A+2sinA}{cosA(1+sinA)}\times (\frac{1-sin^2A-1-sin^2A+2sinA}{cosA(1-sinA)})$

By using formula

$(a+b)^2=a^2+b^2+2ab$

$(a-b)^2=a^2+b^2-2ab$

$cos^2A=1-sin^2A$

$\frac{1+1+2sinA}{cosA(1+sinA)}(\frac{2sinA-2sin^2A}{cosA(1-sinA})$

By using formula

$sin^2x+cos^2x=1$

$2sinA(\frac{2+2sinA}{cosA(1+sinA)}(\frac{1-sinA}{cosA(1-sinA)}))$

$\frac{4sinA(1+sinA)}{cosA(1+sinA)}\times \frac{1-sinA}{cosA(1-sinA)}$

$\frac{4sinA}{cosA\cdot cosA}$

$4tanAsecA$=RHS

By using formula

$\frac{sinA}{cosA}=tanA$

$\frac{1}{cosA}=secA$

Hence, proved.

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