If tan^2⁡x+y+cot^2⁡x+y=1-2x-x^2, then where, n∈Z

Correct answer: (B) x=-1,y=nπ±(π)/4+1. Left hand side is greater than or equal to 2. ⇒1-2x-x^2≥2 ⇒x+1^2≤0 ⇒x=-1 Hence, the given equation is tan^2⁡y-1+cot^2⁡y-1=2 ⇒tan⁡y-1=±1 ⇒y-1=nπ±(π)/4 ⇒y=nπ±(π)/4+1 So, x=-1 and y=nπ±(π)/4+1

Q32 / 83 · NTA Abhyas (JEE Main) 2020 · Mathematics · Trigonometric Equations

If $\tan^{2} (x + y) + \cot^{2} (x + y) = 1 - 2 x - x^{2},$ then $(w h e r e, n \in Z)$

A
$x = 1, y = n π \pm \frac{π}{4} - 1$
Correct answer
B
$x = - 1, y = n π \pm \frac{π}{4} + 1$
C
$x = - 1, y = \frac{n π}{2} \pm \frac{π}{4} - 1$
D
$x = + 1, y = \frac{n π}{2} \pm \frac{π}{4} - 1$

Solution

Left hand side is greater than or equal to

2 .

\Rightarrow 1 - 2 x - x^{2} \geq 2

\Rightarrow {(x + 1)}^{2} \leq 0

\Rightarrow x = - 1

Hence, the given equation is

\tan^{2} (y - 1) + \cot^{2} (y - 1) = 2

\Rightarrow \tan (y - 1) = \pm 1

\Rightarrow y - 1 = nπ \pm \frac{π}{4}

\Rightarrow y = (nπ \pm \frac{π}{4}) + 1

So,

x = - 1

and

y = (n π \pm \frac{π}{4}) + 1

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

If tan2⁡x+y+cot2⁡x+y=1-2x-x2, then where, n∈Z

Solution

More Trigonometric Equations Questions

If $\tan^{2} (x + y) + \cot^{2} (x + y) = 1 - 2 x - x^{2},$ then $(w h e r e, n \in Z)$