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Question

2tan1[aba+btanθ2]=

A
cos1(acosθ+babcosθ)
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B
cos1(acosθbabcosθ)
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C
cos1(a+bcosθacosθ+b)
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D
cos1(abcosθacosθb)
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Solution

The correct option is A cos1(acosθ+babcosθ)

We have,

2tan1[aba+btanθ2]=?

Solve:-

We know that,

2tan1x=cos1(1x21+x2)

Then,

We can use this formula,

2tan1[aba+btanθ2]=cos1⎢ ⎢ ⎢ ⎢ ⎢1(aba+btanθ2)21+(aba+btanθ2)2⎥ ⎥ ⎥ ⎥ ⎥

=cos1⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢1aba+bsin2θ2cos2θ21+aba+bsin2θ2cos2θ2⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥

=cos1⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢(a+b)cos2θ2(ab)sin2θ2(a+b)cos2θ2(a+b)cos2θ2+(ab)sin2θ2(a+b)cos2θ2⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥

=cos1⎢ ⎢ ⎢(a+b)cos2θ2(ab)sin2θ2(a+b)cos2θ2+(ab)sin2θ2⎥ ⎥ ⎥

=cos1⎢ ⎢ ⎢acos2θ2+bcos2θ2asin2θ2+bsin2θ2acos2θ2+bcos2θ2+asin2θ2bsin2θ2⎥ ⎥ ⎥

=cos1⎢ ⎢ ⎢acos2θ2asin2θ2+bcos2θ2+bsin2θ2acos2θ2+asin2θ2bsin2θ2+bcos2θ2⎥ ⎥ ⎥

=cos1⎢ ⎢ ⎢ ⎢a(cos2θ2sin2θ2)+b(cos2θ2+sin2θ2)a(cos2θ2+sin2θ2)b(bcos2θ2sin2θ2)⎥ ⎥ ⎥ ⎥

=cos1[acosθ+b×1a×1bcosθ]

=cos1[acosθ+babcosθ]

Hence, this is the answer.

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