If sin⁡x+cos⁡x=(√7)/2, where x∈0,(π)/4, then the value of tanx/2 is equal to

Correct answer: (B) (√7-2)/3. sin⁡x+cos⁡x=(√7)/2 ⇒2tanx2/(1+tan^2x2)+(1-tan^2x2)/(1+tan^2x2)=√72 ⇒4tanx/2+2-2tan^2x/2=√7+√7tan^2x/2 ⇒2+√7tan^2x/2-4tanx/2+√7-2=0 ⇒tanx/2=(4±√(16-42+7)√7-2)/(22+√7) =(4±√(16-4×3))/(22+√7) tanx/2=(2±1)/(2+√7)=1/(2+√7) or 3/(2+√7) tanx/2=(√7-2)/3 or √7-2 But, as x/2∈0,(π)/8 ⇒tanx/2=(√7-2)/3

Q46 / 83 · NTA Abhyas (JEE Main) 2020 · Mathematics · Trigonometric Equations

If $\sin x + \cos x = \frac{\sqrt{7}}{2}$ , where $x \in [0, \frac{π}{4}]$ , then the value of $t a n \frac{x}{2}$ is equal to

A
$\frac{3 - \sqrt{7}}{3}$
Correct answer
B
$\frac{\sqrt{7} - 2}{3}$
C
$\frac{4 - \sqrt{7}}{4}$
D
$\frac{5 - \sqrt{3}}{2}$

Solution

\sin x + \cos x = \frac{\sqrt{7}}{2}

\Rightarrow \frac{2 t a n \frac{x}{2}}{1 + t a n^{2} \frac{x}{2}} + \frac{1 - t a n^{2} \frac{x}{2}}{1 + t a n^{2} \frac{x}{2}} = \frac{\sqrt{7}}{2}

\Rightarrow 4 t a n \frac{x}{2} + 2 - 2 t a n^{2} \frac{x}{2} = \sqrt{7} + \sqrt{7} t a n^{2} \frac{x}{2}

\Rightarrow (2 + \sqrt{7}) t a n^{2} \frac{x}{2} - 4 t a n \frac{x}{2} + \sqrt{7} - 2 = 0

\Rightarrow t a n \frac{x}{2} = \frac{4 \pm \sqrt{16 - 4 (2 + \sqrt{7}) (\sqrt{7} - 2)}}{2 (2 + \sqrt{7})}

= \frac{4 \pm \sqrt{16 - 4 \times 3}}{2 (2 + \sqrt{7})}

\tan \frac{x}{2} = \frac{2 \pm 1}{2 + \sqrt{7}} = \frac{1}{2 + \sqrt{7}} or \frac{3}{2 + \sqrt{7}}

t a n \frac{x}{2} = \frac{\sqrt{7} - 2}{3}

\sqrt{7} - 2

But, as

\frac{x}{2} \in [0, \frac{π}{8}]

\Rightarrow t a n \frac{x}{2} = \frac{\sqrt{7} - 2}{3}

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

If sin⁡x+cos⁡x=72, where x∈0,π4, then the value of tanx2 is equal to

Solution

More Trigonometric Equations Questions

If $\sin x + \cos x = \frac{\sqrt{7}}{2}$ , where $x \in [0, \frac{π}{4}]$ , then the value of $t a n \frac{x}{2}$ is equal to