If 0 - A -4- then tan -11-2tan 2 A -tan -1 A -tan -1 3 A is equal toA- -4- 0- -D- -2

The correct option is C πtan^ 1 (12tan 2A) + tan^ 1 ( A) + tan^ 1 (^3 A) = tan^ 1(12tan 2A) + tan^ 1( A + ^3A1 ^4 A) = tan^ 1(12·2tan A1 tan^2 A) + tan^ 1( A ...

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Byju's Answer

Standard XII

Mathematics

Integration of Trigonometric Functions

If 0 ≤ A ≤π/4...

Question

If $0 \leq A \leq \frac{π}{4},$ then ${tan}^{- 1} (\frac{1}{2} tan 2 A) + {tan}^{- 1} (cot A) + {tan}^{- 1} ({cot}^{3} A)$ is equal to

\frac{π}{4}

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π

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0

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\frac{π}{2}

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Solution

The correct option is B $π$
${tan}^{- 1} (\frac{1}{2} tan 2 A) + {tan}^{- 1} (cot A) + {tan}^{- 1} ({cot}^{3} A) = {tan}^{- 1} (\frac{1}{2} tan 2 A) + {tan}^{- 1} (\frac{cot A + {cot}^{3} A}{1 - {cot}^{4} A}) = {tan}^{- 1} (\frac{1}{2} \cdot \frac{2 tan A}{1 - {tan}^{2} A}) + {tan}^{- 1} (\frac{cot A (1 + {cot}^{2} A)}{(1 - {cot}^{2} A) (1 + {cot}^{2} A)}) = {tan}^{- 1} (\frac{tan A}{1 - {tan}^{2} A}) + {tan}^{- 1} (\frac{tan A}{{tan}^{2} A - 1}) = {tan}^{- 1} (\frac{tan A}{1 - {tan}^{2} A}) + {tan}^{- 1} (\frac{- tan A}{1 - {tan}^{2} A}) = {tan}^{- 1} (\frac{tan A}{1 - {tan}^{2} A}) + π - {tan}^{- 1} (\frac{tan A}{1 - {tan}^{2} A}) = π + 0 = π$

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Similar questions

Q. Prove that:

{tan}^{- 1} (\frac{1}{2} tan 2 A) + {tan}^{- 1} (cot A) + {tan}^{- 1} ({cot}^{3} A) = ⎧ ⎨ ⎩ \begin{matrix} 0, & i f \frac{π}{4} < A < \frac{π}{2} π, & i f 0 < A < π / 4 \end{matrix}

Q. The value of

{tan}^{- 1} (\frac{1}{2} tan 2 A) + {tan}^{- 1} (cot A) + {tan}^{- 1} ({cot}^{3} A)

for

0 < A < \frac{π}{4}

is?

Q. Find the value of

{tan}^{- 1} (\frac{1}{2} tan 2 A) + {tan}^{- 1} (cot A) + {tan}^{- 1} ({cot}^{3} A)

, for

0 < A < \frac{π}{4}

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

.
Prove that

{t a n}^{- 1} (\frac{1}{2} t a n 2 A) + {t a n}^{- 1} (c o t A) + {t a n}^{- 1} ({c o t}^{3} A)

0

π / 4 < A < π / 2

and

= π

0 < A < π / 4

Q. The value of

{tan}^{- 1} (\frac{1}{2} tan 2 A) + {tan}^{- 1} (c o t A) + t a n^{- 1} (c o t^{3} A)

for

0 < A < π / 4

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If 0≤A≤π4, then tan−1(12tan2A)+tan−1(cotA)+tan−1(cot3A) is equal to

If $0 \leq A \leq \frac{π}{4},$ then ${tan}^{- 1} (\frac{1}{2} tan 2 A) + {tan}^{- 1} (cot A) + {tan}^{- 1} ({cot}^{3} A)$ is equal to