If cotα=1 and secβ=-5/3, where π<α<(3π)/2 and (π)/2<β<π, then the value of tanα+β and the quadrant in which α+β lies, respectively are

Correct answer: (A) -1/7 and IV^th quadrant. Given, cotα=1,secβ=(-5)/3 So, cosβ=(-3)/5,tanβ=(-4)/3 and tanα=1 Now using formula tanα+β=(tanα+tanβ)/(1-tanαtanβ) ⇒tanα+β=(1-43)/(1+43×1)=(-1)/7 We know that tanθ is negative in 2^nd & 4^th quadrant but as π<α<(3π)/2, so α+β will lie in IV^th quadrant.

Q45 / 90 · JEE Main 2022 (28 Jun Shift 2) · Mathematics

If $\cot α = 1$ and $\sec β = - \frac{5}{3}$ , where $π < α < \frac{3 π}{2}$ and $\frac{π}{2} < β < π$ , then the value of $\tan (α + β)$ and the quadrant in which $α + β$ lies, respectively are

Correct answer
A
$- \frac{1}{7}$ and ${IV}^{th}$ quadrant
B
$7$ and $I^{st}$ quadrant
C
$- 7$ and ${IV}^{th}$ quadrant
D
$\frac{1}{7}$ and $I^{st}$ quadrant

Solution

Given,

$\cot α = 1, \sec β = \frac{- 5}{3}$

So, $\cos β = \frac{- 3}{5}, \tan β = \frac{- 4}{3}$ and $\tan α = 1$

Now using formula $\tan (α + β) = \frac{\tan α + \tan β}{1 - \tan α \tan β}$

$\Rightarrow \tan (α + β) = \frac{1 - \frac{4}{3}}{1 + \frac{4}{3} \times 1} = \frac{- 1}{7}$

We know that $\tan θ$ is negative in $2^{n d} & 4^{t h}$ quadrant but as $π < α < \frac{3 π}{2}$ , so $α + β$ will lie in ${IV}^{th}$ quadrant.

Hence, the correct answer is option: 'A'.

If cotα=1 and secβ=-53, where π<α<3π2 and π2<β<π, then the value of tanα+β and the quadrant in which α+β lies, respectively are

Solution

Relevant Questions

If $\cot α = 1$ and $\sec β = - \frac{5}{3}$ , where $π < α < \frac{3 π}{2}$ and $\frac{π}{2} < β < π$ , then the value of $\tan (α + β)$ and the quadrant in which $α + β$ lies, respectively are