2tan−1[√a−ba+btanθ2]=
The correct option is A cos−1(acosθ+ba−bcosθ)
We have,
2tan−1[√a−ba+btanθ2]=?
Solve:-
We know that,
2tan−1x=cos−1(1−x21+x2)
Then,
We can use this formula,
2tan−1[√a−ba+btanθ2]=cos−1⎡⎢
⎢
⎢
⎢
⎢⎣1−(√a−ba+btanθ2)21+(√a−ba+btanθ2)2⎤⎥
⎥
⎥
⎥
⎥⎦
=cos−1⎡⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢⎣1−a−ba+bsin2θ2cos2θ21+a−ba+bsin2θ2cos2θ2⎤⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥⎦
=cos−1⎡⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢⎣(a+b)cos2θ2−(a−b)sin2θ2(a+b)cos2θ2(a+b)cos2θ2+(a−b)sin2θ2(a+b)cos2θ2⎤⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥⎦
=cos−1⎡⎢
⎢
⎢⎣(a+b)cos2θ2−(a−b)sin2θ2(a+b)cos2θ2+(a−b)sin2θ2⎤⎥
⎥
⎥⎦
=cos−1⎡⎢
⎢
⎢⎣acos2θ2+bcos2θ2−asin2θ2+bsin2θ2acos2θ2+bcos2θ2+asin2θ2−bsin2θ2⎤⎥
⎥
⎥⎦
=cos−1⎡⎢
⎢
⎢⎣acos2θ2−asin2θ2+bcos2θ2+bsin2θ2acos2θ2+asin2θ2−bsin2θ2+bcos2θ2⎤⎥
⎥
⎥⎦
=cos−1⎡⎢
⎢
⎢
⎢⎣a(cos2θ2−sin2θ2)+b(cos2θ2+sin2θ2)a(cos2θ2+sin2θ2)−b(bcos2θ2−sin2θ2)⎤⎥
⎥
⎥
⎥⎦
=cos−1[acosθ+b×1a×1−bcosθ]
=cos−1[acosθ+ba−bcosθ]
Hence, this is the
answer.
We have,
2tan−1[√a−ba+btanθ2]=?
Solve:-
We know that,
2tan−1x=cos−1(1−x21+x2)
Then,
We can use this formula,
2tan−1[√a−ba+btanθ2]=cos−1⎡⎢ ⎢ ⎢ ⎢ ⎢⎣1−(√a−ba+btanθ2)21+(√a−ba+btanθ2)2⎤⎥ ⎥ ⎥ ⎥ ⎥⎦
=cos−1⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣1−a−ba+bsin2θ2cos2θ21+a−ba+bsin2θ2cos2θ2⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦
=cos−1⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣(a+b)cos2θ2−(a−b)sin2θ2(a+b)cos2θ2(a+b)cos2θ2+(a−b)sin2θ2(a+b)cos2θ2⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦
=cos−1⎡⎢ ⎢ ⎢⎣(a+b)cos2θ2−(a−b)sin2θ2(a+b)cos2θ2+(a−b)sin2θ2⎤⎥ ⎥ ⎥⎦
=cos−1⎡⎢ ⎢ ⎢⎣acos2θ2+bcos2θ2−asin2θ2+bsin2θ2acos2θ2+bcos2θ2+asin2θ2−bsin2θ2⎤⎥ ⎥ ⎥⎦
=cos−1⎡⎢ ⎢ ⎢⎣acos2θ2−asin2θ2+bcos2θ2+bsin2θ2acos2θ2+asin2θ2−bsin2θ2+bcos2θ2⎤⎥ ⎥ ⎥⎦
=cos−1⎡⎢ ⎢ ⎢ ⎢⎣a(cos2θ2−sin2θ2)+b(cos2θ2+sin2θ2)a(cos2θ2+sin2θ2)−b(bcos2θ2−sin2θ2)⎤⎥ ⎥ ⎥ ⎥⎦
=cos−1[acosθ+b×1a×1−bcosθ]
=cos−1[acosθ+ba−bcosθ]
Hence, this is the answer.
