Tan^ 1 ( tan3π/4 )=tan^ 1 [ tan ( π π/4 ) ]=tan^ 1 ( tanπ/4 ) [∵ tan(π θ)= tanθ] =tan^ 1 [ tan ( π/4 ) ] [∵ tanθ=tan( θ)] = π/4 which lies in the r ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios for Sum of Two Angles

tan -1tan3 π/...

Question

$t a n^{- 1} (t a n \frac{3 π}{4})$

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Solution

$t a n^{- 1} (t a n \frac{3 π}{4}) = t a n^{- 1} [t a n (π - \frac{π}{4})] = t a n^{- 1} (- t a n \frac{π}{4}) [∵ t a n (π - θ) = - t a n θ] = t a n^{- 1} [t a n (- \frac{π}{4})] [∵ - t a n θ = t a n (- θ)]$
$= - \frac{π}{4}$ which lies in the range $(- \frac{π}{2}, \frac{π}{2})$

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Similar questions

tan^{- 1} (tan 3 π / 4) =

The value of

{tan}^{- 1} (tan \frac{3 π}{4})

is?

Q. If the value of

{tan}^{- 1} (tan \frac{3 π}{4})

- \frac{π}{k},

then

k

Q. Inverse circular functions,Principal values of

{s i n}^{- 1} x, {c o s}^{- 1} x, {t a n}^{- 1} x

{t a n}^{- 1} x + {t a n}^{- 1} y = {t a n}^{- 1} \frac{x + y}{1 - x y}

x y < 1

π + {t a n}^{- 1} \frac{x + y}{1 - x y}

x y > 1

.
(a)

{t a n}^{- 1} (t a n \frac{3 π}{4})

(b)

{t a n}^{- 1} (t a n \frac{2 π}{3})

(c)

c o s [{c o s}^{- 1} (\frac{- \sqrt{3}}{2}) = \frac{π}{6}]

(d) If

a \leq {t a n}^{- 1} (\frac{1 - x}{1 + x}) \leq b

where

0 \leq x \leq 1

tan θ + tan 4 θ + tan 7 θ = tan θ tan 4 θ tan 7 θ

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tan−1(tan3π4)

tan−1(tan3π4)=tan−1[tan(π−π4)]=tan−1(−tanπ4) [∵tan(π−θ)=−tanθ]=tan−1[tan(−π4)] [∵−tanθ=tan(−θ)] =−π4 which lies in the range (−π2,π2)

$t a n^{- 1} (t a n \frac{3 π}{4})$

$t a n^{- 1} (t a n \frac{3 π}{4}) = t a n^{- 1} [t a n (π - \frac{π}{4})] = t a n^{- 1} (- t a n \frac{π}{4}) [∵ t a n (π - θ) = - t a n θ] = t a n^{- 1} [t a n (- \frac{π}{4})] [∵ - t a n θ = t a n (- θ)]$
$= - \frac{π}{4}$ which lies in the range $(- \frac{π}{2}, \frac{π}{2})$