If tanA + cotA = 4 , then tan^4A + cot^4A is equal to

English

Gujarati

Hindi

Trigonometrical Equations

normal

If $tanA + cotA = 4$, then $tan^4A + cot^4A$ is equal to

A

$110$

B

$191$

C

$80$

D

$194$

Solution

$(\tan A+\cot A)^{2}=16$

$\Rightarrow \quad \tan ^{2} A+\cot ^{2} A+2=16$

$\Rightarrow \quad \tan ^{2} A+\cot ^{2} A=14$

$\Rightarrow \quad\left(\tan ^{2} A+\cot ^{2} A\right)^{2}=196$

$\Rightarrow \quad \tan ^{4} A+\cot ^{4} A+2=196$

$\Rightarrow \quad \tan ^{4} A+\cot ^{4} A=194$

Standard 11

Mathematics

Share

Similar Questions

The number of elements in the set $S=$ $\left\{\theta \in[-4 \pi, 4 \pi]: 3 \cos ^{2} 2 \theta+6 \cos 2 \theta-\right.$ $\left.10 \cos ^{2} \theta+5=0\right\}$ is

hard

(JEE MAIN-2022)

$\cot \theta = \sin 2\theta (\theta \ne n\pi $, $n$ is integer), if $\theta = $

easy

Number of solution $(s)$ of the equation ${\cos ^2}2x + {\cos ^2}\frac{{5x}}{4} = \cos 2x\,{\cos ^2}5x$ in $\left[ {0,\frac{\pi }{3}} \right]$ is

normal

Find the general solution of the equation $\cos 3 x+\cos x-\cos 2 x=0$

medium

If equation in variable $\theta, 3 tan(\theta -\alpha) = tan(\theta + \alpha)$, (where $\alpha$ is constant) has no real solution, then $\alpha$ can be (wherever $tan(\theta – \alpha)$ & $tan(\theta + \alpha)$ both are defined)

normal