Eliminate - between a-cos- - and b-sin- -

Solution:(θ α )=cos^ 1(a)—–(1)0ex0ex(θ β )=sin^ 1(b)—–(2)0ex0exsubtracting the equation, we get0ex0ex(β α )=cos^ 1(a) sin^ 1(b)Hence, (β α )=cos^ 1(a) sin^ ...

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Standard XII

Mathematics

Angle between Two Vectors

Eliminate θ b...

Question

Eliminate $θ$ between $a = \cos$ ( $θ - α$ ) and $b = \sin$ ( $θ - β$ )

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Solution

Solution:
$(θ - α) = \cos^{- 1} (a) - - - - - (1) (θ - β) = \sin^{- 1} (b) - - - - - (2) s u b t r a c t i n g t h e e q u a t i o n, w e g e t (β - α) = \cos^{- 1} (a) - \sin^{- 1} (b)$
Hence, $(β - α) = \cos^{- 1} (a) - \sin^{- 1} (b)$ where $θ$ is eliminated.

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Eliminate θ between a=cos(θ-α) and b=sin(θ-β)

Solution:(θ-α)=cos-1(a)-----(1)(θ-β)=sin-1(b)-----(2)subtractingtheequation,weget(β-α)=cos-1(a)-sin-1(b)Hence, (β-α)=cos-1(a)-sin-1(b) where θ is eliminated.

Eliminate $θ$ between $a = \cos$ ( $θ - α$ ) and $b = \sin$ ( $θ - β$ )

Solution:
$(θ - α) = \cos^{- 1} (a) - - - - - (1) (θ - β) = \sin^{- 1} (b) - - - - - (2) s u b t r a c t i n g t h e e q u a t i o n, w e g e t (β - α) = \cos^{- 1} (a) - \sin^{- 1} (b)$
Hence, $(β - α) = \cos^{- 1} (a) - \sin^{- 1} (b)$ where $θ$ is eliminated.