2 tan -1cos-tan -12 cos- then- show that- -4

Here, 2tan^ 1 (cos x) = tan^ 1 (2 cosec x) Or tan^ 1 (2cos x/1 cos^2x) = tan^ 1 (2 cosec x) [use 2 tan^ 1 x = tan^ 1 [2x/1 x^2 ]] Or 2 cos x/sin^2 x ...

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Byju's Answer

Standard XII

Mathematics

Composition of Inverse Trigonometric Functions and Trigonometric Functions

2 tan -1cosθ=...

Question

$2 t a n^{- 1} (c o s θ) = t a n^{- 1} (2 c o s s e c θ)$ then, show that, $θ = \frac{π}{4}$

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Solution

Here, $2 t a n^{- 1} (c o s x) = t a n^{- 1} (2 c o s e c x)$
Or $t a n^{- 1} (\frac{2 c o s x}{1} - c o s^{2} x) = t a n^{- 1} (2 c o s e c x) [u s e 2 t a n^{- 1} x = t a n^{- 1} [\frac{2 x}{1} - x^{2}]]$
Or $2 \frac{c o s x}{s i n^{2} x} = \frac{2}{s i n x}$
Or $s i n x = c o s x$
Or $t a n x = 1$
Or $x = \frac{π}{4}$ .

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Similar questions

If 2 $t a n^{- 1} (c o s θ) = t a n^{- 1} (2 c o s e c θ), then show that θ = \frac{π}{4},$ where $θ$ is any integer.

Show that 2 $t a n^{- 1} (- 3) = \frac{- π}{2} + t a n^{- 1} (\frac{- 4}{3}) .$

Q. If

tan [\frac{π}{4} + θ] + tan [\frac{π}{4} - θ] = α

then

{tan}^{3} [\frac{π}{4} + θ] + {tan}^{3} [\frac{π}{4} - θ] =

Q. If

(1 - tan θ) (1 + tan θ) {sec}^{2} θ + 2^{{tan}^{2} θ} = 0

, then the number of values of

θ

in the interval

(- \frac{π}{2}, \frac{π}{2})

Q. The general solution of

(\sqrt{3} - 1) sin θ + (\sqrt{3} + 1) cos θ = 2

is
(where

n \in Z

)

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2tan−1(cosθ)=tan−1(2 cos sec θ) then, show that, θ=π4

Here, 2tan−1(cosx)=tan−1(2cosec x) Or tan−1(2cos x1−cos2x)=tan−1(2cosec x) [use 2 tan−1x=tan−1[2x1−x2]] Or 2cosxsin2x=2sin x Or sinx=cosx Or tanx=1 Or x=π4.

$2 t a n^{- 1} (c o s θ) = t a n^{- 1} (2 c o s s e c θ)$ then, show that, $θ = \frac{π}{4}$