Tan -1-3 sin 2 -5-3cos 2 -tan -1-tan-4-4 where -2-2- then - is -

The correct option is D 4 tan^ 1 [ 3sin2α/5+3cos2α ]+tan^ 1 [ tanα/4 ] = tan^ 1 ( 6tan α/8+2tan^2α )+tan^ 1 ( tanα/4 ) (∵3tan^2α/16+4tan^2α lt;1 ) =tan^ ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Equations

tan -1[3 sin ...

Question

$t a n^{- 1} [\frac{3 s i n 2 α}{5 + 3 c o s 2 α}] + t a n^{- 1} [\frac{t a n α}{4}] = \frac{λ α}{4}$ where $- \frac{π}{2} < α < \frac{π}{2}, t h e n λ$ is ___.

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Solution

The correct option is D
4

$t a n^{- 1} [\frac{3 s i n 2 α}{5 + 3 c o s 2 α}] + t a n^{- 1} [\frac{t a n α}{4}] = t a n^{- 1} (\frac{6 t a n α}{8 + 2 t a n^{2} α}) + t a n^{- 1} (\frac{t a n α}{4}) (∵ \frac{3 t a n^{2} α}{16 + 4 t a n^{2} α} < 1) = t a n^{- 1} (t a n α) = α$

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Similar questions

{tan}^{- 1} (\frac{3 sin 2 α}{5 + 3 cos 2 α}) + {tan}^{- 1} (\frac{tan α}{4}) = α

(where

- \frac{π}{2} < α < \frac{π}{2}

)

Q. If x=

{tan}^{- 1} [\frac{3 sin 2 α}{5 + 3 cos 2 α}] + {tan}^{- 1} [\frac{tan α}{4}]

, where

- \frac{π}{2} < α < \frac{π}{2}

.
Find the value of x.

Q. Solve:

{tan}^{- 1} [\frac{3 sin 2 α}{5 + 3 cos 2 α}] + {tan}^{- 1} (\frac{1}{4} tan α)

Q. If

u = c o t^{- 1} \sqrt{t a n α} - t a n^{- 1} \sqrt{t a n α}, t h e n t a n (\frac{π}{4} - \frac{u}{2})

is equal to

Q. Show that

2 {tan}^{- 1} {tan \frac{α}{2} tan (\frac{π}{4} - \frac{β}{2})}

= {tan}^{- 1} (\frac{sin α cos β}{cos α + sin β})

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tan−1[3sin2α5+3cos2α]+tan−1[tanα4]=λα4 where −π2<α<π2, then λ is ___.

The correct option is D 4 tan−1[3sin2α5+3cos2α]+tan−1[tanα4]=tan−1(6tanα8+2tan2α)+tan−1(tanα4)(∵3tan2α16+4tan2α<1)=tan−1(tanα)=α

$t a n^{- 1} [\frac{3 s i n 2 α}{5 + 3 c o s 2 α}] + t a n^{- 1} [\frac{t a n α}{4}] = \frac{λ α}{4}$ where $- \frac{π}{2} < α < \frac{π}{2}, t h e n λ$ is ___.

The correct option is D
4

$t a n^{- 1} [\frac{3 s i n 2 α}{5 + 3 c o s 2 α}] + t a n^{- 1} [\frac{t a n α}{4}] = t a n^{- 1} (\frac{6 t a n α}{8 + 2 t a n^{2} α}) + t a n^{- 1} (\frac{t a n α}{4}) (∵ \frac{3 t a n^{2} α}{16 + 4 t a n^{2} α} < 1) = t a n^{- 1} (t a n α) = α$