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Find the value of : (2{}^{-1}{1}{4} )+ ( {}^{-1} 2 2 )

Question

Find the value of : $sin (2 tan^{- 1} \frac{1}{4}) + cos (tan^{- 1} 22)$

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Find the value of :

sin (2 tan^{- 1} \frac{1}{4}) + cos (tan^{- 1} 22)

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Text solutionVerified

Consider

sin (2 tan^{- 1} \frac{1}{4})

Let

tan^{- 1} \frac{1}{4} = θ

\Rightarrow tan θ = \frac{1}{4}

.......

(1)

\Rightarrow sin (2 tan^{- 1} \frac{1}{4}) = sin 2 θ

= \frac{2 tan θ}{1 + tan ^{2} θ}

= \frac{2 \times \frac{1}{4}}{1 + ( \frac{1}{4} ) ^{2}}

= \frac{\frac{1}{2}}{\frac{16 + 1}{16}} = \frac{1}{2} \times \frac{16}{17} = \frac{8}{17}

∴ sin (2 tan^{- 1} \frac{1}{4}) = \frac{8}{17}

Now consider

cos (tan^{- 1} 22)

Let

tan^{- 1} 22 = ϕ

∴ tan ϕ = 22

\Rightarrow cos (tan^{- 1} 22) = cos ϕ

We know that

1 + tan^{2} ϕ = sec^{2} ϕ

\Rightarrow 1 + (22)^{2} = sec^{2} ϕ

\Rightarrow sec^{2} ϕ = 1 + 8 = 9

\Rightarrow \frac{1}{cos ^{2} ϕ} = 9

\Rightarrow cos ϕ = \pm \frac{1}{3}

Since value of

tan ϕ

is positive, we take positive value of

cos ϕ

∴ cos ϕ = \frac{1}{3}

So,

cos (tan^{- 1} 22) = \frac{1}{3}

∴ sin (2 tan^{- 1} \frac{1}{4}) + cos (tan^{- 1} 22)

= \frac{8}{17} + \frac{1}{3} = \frac{24 + 17}{51} = \frac{41}{51}

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Question Text	Find the value of : $sin (2 tan^{- 1} \frac{1}{4}) + cos (tan^{- 1} 22)$
Answer Type	Text solution:1
Upvotes	150