Math, asked by mezank62, 6 months ago

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Answers

Answered by Abhisheksingh5722
6

Answer:

Answer:

\LARGE{\bf{\underline{\underline{GIVEN:-}}}}

GIVEN:−

\sf \bullet \ \ \dfrac{(1+sinA-cosA)^2}{(1+sinA+cosA)^2} < /p > < p >∙

(1+sinA+cosA)

2

(1+sinA−cosA)

2

</p><p>

\LARGE{\bf{\underline{\underline{SOLUTION:-}}}}

SOLUTION:−

LHS:

\sf \to \dfrac{(1+sinA-cosA)^2}{(1+sinA+cosA)^2}→

(1+sinA+cosA)

2

(1+sinA−cosA)

2

Expand the fractions using .

\sf \to \dfrac{(cos^2-2sincos+sin^2-2cos+2sin+1)}{(cos^2+2sincos+sin^2+2cos+2sin+1)}→

(cos

2

+2sincos+sin

2

+2cos+2sin+1)

(cos

2

−2sincos+sin

2

−2cos+2sin+1)

Rearrange the terms.

\sf \to \dfrac{(cos^2+sin^2-2sincos-2cos+2sin+1)}{(cos^2+sin^2+2sincos+2cos+2sin+1)}→

(cos

2

+sin

2

+2sincos+2cos+2sin+1)

(cos

2

+sin

2

−2sincos−2cos+2sin+1)

We know that cos²A+sin²A=1.

\sf \to \dfrac{1-2sincos-2cos}{2sin+1}→

2sin+1

1−2sincos−2cos

Now here, take -2cos common from the numerator and +2cos common from the denominator.

\sf \to \dfrac{1-2cos(sin+2)}{2sin+1}→

2sin+1

1−2cos(sin+2)

Now, rearrange the terms, add 1 and 1 and take 2 common.

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