If tan -1 x-tan -1 y-4 -5- then -1 x- -1 y is equal toA- -B- 2 -5- 3 -5D-

The correct option is D π/5Given that nbsp;tan^ 1x+tan^ 1y=4π/5.....(1) Let ^ 1x+^ 1y = p.....(2) we know that tan^ 1a+^ 1a=π/2 Adding (1) and (2) tan^ 1x+^ ...

You visited us 4 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios for Sum of Two Angles

If tan -1 x+t...

Question

$If {tan}^{- 1} x + {tan}^{- 1} y = \frac{4 π}{5}, then {cot}^{- 1} x + {cot}^{- 1} y$ $is equal to$

π

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{π}{5}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{2 π}{5}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{3 π}{5}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $\frac{π}{5}$
Given that ${tan}^{- 1} x + {tan}^{- 1} y = \frac{4 π}{5}$ .....(1)
Let ${cot}^{- 1} x + {cot}^{- 1} y = p$ .....(2)
we know that ${tan}^{- 1} a + {cot}^{- 1} a = \frac{π}{2}$

Adding (1) and (2)
${tan}^{- 1} x + {cot}^{- 1} x + {tan}^{- 1} y + {cot}^{- 1} y = \frac{4 π}{5} + p$
$\Rightarrow \frac{π}{2} + \frac{π}{2} = \frac{4 π}{5} + p$
$\Rightarrow p = \frac{π}{5}$

Suggest Corrections

Similar questions

If $t a n^{- 1} x + t a n^{- 1} y = \frac{4 π}{5}, t h e n c o t^{- 1} x + c o t^{- 1} y$ equals to

(a) $\frac{π}{5}$ (b) $\frac{2 π}{5}$ (c) $\frac{3 π}{5}$ (d) $π$

Q. If

t a n^{- 1} x + t a n^{- 1} y = \frac{4 π}{5}

, then

c o t^{- 1} x + c o t^{- 1} y

is equal to

Q. If tan^-1x + tan^-1 y =

\frac{5 π}{6}

, then cot^-1x + cot^-1y = _________________.

If $s i n^{- 1} x + s i n^{- 1} y = \frac{π}{2}$ ,then $c o s^{- 1} x + c o s^{- 1} y$ is equal to

If ${cos}^{- 1} x + {cos}^{- 1} y = \frac{π}{2}$ and ${tan}^{- 1} x - {tan}^{- 1} y = 0$ then $x^{2} + x y + y^{2}$ is equal to

Join BYJU'S Learning Program

If tan−1x+tan−1y=4π5, then cot−1x+cot−1y is equal to

The correct option is B π5Given that tan−1x+tan−1y=4π5.....(1) Let cot−1x+cot−1y=p.....(2) we know that tan−1a+cot−1a=π2 Adding (1) and (2) tan−1x+cot−1x+tan−1y+cot−1y=4π5+p ⇒π2+π2=4π5+p ⇒p=π5

$If {tan}^{- 1} x + {tan}^{- 1} y = \frac{4 π}{5}, then {cot}^{- 1} x + {cot}^{- 1} y$ $is equal to$