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Find the value of tan22°30′. [Hint: Let θ = 45°, use tan θ2=sin θ2cos θ2=2sin θ2cos θ22cos2 θ2=sinθ1+cosθ]

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Question

Find the value of tan22°30′. `["Hint:"  "Let" θ = 45°, "use" tan  theta/2 = (sin  theta/2)/(cos  theta/2) = (2sin  theta/2 cos  theta/2)/(2cos^2  theta/2) = sintheta/(1 + costheta)]`

Sum
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Solution

Let 22°30′ = `theta/2`

∴ θ = 45°

tan22°30′ = `tan  theta/2`

= `(sin  theta/2)/(cos  theta/2)`

= `(2sin  theta/2 cos  theta/2)/(2cos^2  theta/2)`

= `sintheta/(1 + costheta)`

Put θ = 45°

∴ `sintheta/(1 + costheta) = sin 45^circ/(1 + cos 45^circ)`

= `(1/sqrt(2))/(1 + 1/sqrt(2))`

= `1/(sqrt(2) + 1)`

= `(1 xx (sqrt(2) - 1))/((sqrt(2) + 1)(sqrt(2) - 1))`

= `sqrt(2) - 1`

Hence, tan22°30' = `sqrt(2) - 1`.

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Chapter 3: Trigonometric Functions - Exercise [Page 53]

APPEARS IN

NCERT Exemplar Mathematics Exemplar [English] Class 11
Chapter 3 Trigonometric Functions
Exercise | Q 8 | Page 53

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