Tan - 1-2sin-1 2a-1-a2 -1-2cos-1 1-a2-1-a2 -

Tan [ 1/2sin^ 1 ( 2a/1+a^2 ) +1/2cos^ 1 ( 1 a^2/1+a^2 ) ] = tan [ 1/2sin^ 1 ( 2tanθ/1+tan^2θ )+1/2cos^ 1 ( 1 tan^2θ/1+tan^2θ ) ] (Let a=tanθ ) =tan [ 1/2sin^ ...

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Byju's Answer

Standard XII

Mathematics

Properties Derived from Trigonometric Identities

tan [ 1/2sin-...

Question

$t a n [\frac{1}{2} s i n^{- 1} (\frac{2 a}{1 + a^{2}}) + \frac{1}{2} c o s^{- 1} (\frac{1 - a^{2}}{1 + a^{2}})] =$

\frac{2 a}{1 + a^{2}}

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\frac{1 - a^{2}}{1 + a^{2}}

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\frac{2 a}{1 - a^{2}}

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None of these

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Solution

The correct option is C $\frac{2 a}{1 - a^{2}}$
$t a n [\frac{1}{2} s i n^{- 1} (\frac{2 a}{1 + a^{2}}) + \frac{1}{2} c o s^{- 1} (\frac{1 - a^{2}}{1 + a^{2}})]$
$= t a n [\frac{1}{2} s i n^{- 1} (\frac{2 t a n θ}{1 + t a n^{2} θ}) + \frac{1}{2} c o s^{- 1} (\frac{1 - t a n^{2} θ}{1 + t a n^{2} θ})]$
$(L e t a = t a n θ)$
$= t a n [\frac{1}{2} s i n^{- 1} (s i n 2 θ) + \frac{1}{2} c o s^{- 1} (c o s 2 θ)]$
$= t a n (2 θ) = t a n 2 θ = \frac{2 t a n θ}{1 - t a n^{2} θ} = \frac{2 a}{1 - a^{2}}$
Trick : Put a=0, then $t a n (0 + 0) = 0$ ; which is given by (a) and (c). Again put a=1, then $t a n (\frac{π}{4} + \frac{π}{4}) = \infty$ , which is given by (c).

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Similar questions

Inverse circular functions,Principal values of

{sin}^{- 1} x, {c o s}^{- 1} x, {tan}^{- 1} x

{tan}^{- 1} x + {tan}^{- 1} y = {tan}^{- 1} \frac{x + y}{1 - x y}

x y < 1

π + {tan}^{- 1} \frac{x + y}{1 - x y}

x y > 1

(a) If

{sin}^{- 1} \frac{2 p}{1 + p^{2}} - {cos}^{- 1} \frac{1 - q^{2}}{1 + q^{2}} = {tan}^{- 1} \frac{2 x}{1 - x^{2}}

then prove that

x = \frac{p - q}{1 + p q}

(b) Solve for x

{sin}^{- 1} \frac{2 a}{1 + a^{2}} + {sin}^{- 1} \frac{2 b}{1 + b^{2}} = 2 {tan}^{- 1} x

tan [\frac{1}{2} {sin}^{- 1} \frac{2 a}{1 + a^{2}} + \frac{1}{2} {cos}^{- 1} \frac{1 - a^{2}}{1 + a^{2}}] = \frac{2 a}{1 - a^{2}}

If $t a n^{- 1} \frac{1}{a - 1} = t a n^{- 1} \frac{1}{x} + t a n^{- 1} \frac{1}{a^{2} - x + 1}$ , then x is

{sin}^{- 1} \frac{2 a}{1 + a^{2}} + {cos}^{- 1} \frac{1 - b^{2}}{1 + b^{2}} = 2 {tan}^{- 1} x

Q. If

s i n^{- 1} \frac{2 a}{1 + a^{2}} - c o s^{- 1} \frac{1 - b^{2}}{1 + b^{2}} = {t a n}^{- 1} \frac{2 x}{1 - x^{2}}

, then prove that

x = \frac{a - b}{1 + a b}

$(1 + tan A)^{2} + (1 - tan A)^{2} =$

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tan[12sin−1(2a1+a2)+12cos−1(1−a21+a2)]=

$t a n [\frac{1}{2} s i n^{- 1} (\frac{2 a}{1 + a^{2}}) + \frac{1}{2} c o s^{- 1} (\frac{1 - a^{2}}{1 + a^{2}})] =$