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Question

In a triangle ABC right angled at B,
ACB=θ.

If tanθ=512,
then find the value of (1+cos θ)×(1sinθ)(1+sinθ)×(1cosθ).

A
1009
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B
9118
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C
8427
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D
125
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Solution

The correct option is A 1009

ΔABC is right angled at B.
Given,
tanθ=512.Let AB=5k, BC=12k

Applying Pythagoras theorem, we get,

AB2+BC2=AC2

AC2=(12k)2+(5k)2

AC2=169k2

AC=13k

So, sinθ=ABAC=513

and cosθ=BCAC=1213

Hence,

(1+cos θ)×(1sin θ)(1+sin θ)×(1cos θ)=(1+1213)×(1513)(1+513)×(11213)

=2513×8131813×113

=20018=1009

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