(C) Given the equation $2 \tan x - k (1 + \tan^2 x) = 0$,we can rewrite it as $k = \frac{2 \tan x}{1 + \tan^2 x}$.Using the identity $\sin 2x = \frac{

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(C) Given the equation $2 \tan x - k (1 + \tan^2 x) = 0$,we can rewrite it as $k = \frac{2 \tan x}{1 + \tan^2 x}$.Using the identity $\sin 2x = \frac{2 \tan x}{1 + \tan^2 x}$,we get $\sin 2x = k$.Since $B$ and $C$ satisfy this equation,$\sin 2B = k$ and $\sin 2C = k$,so $\sin 2B = \sin 2C$.This implies $2C = 2B$ (not possible as $B Thus,$2B + 2C = \pi$,which simplifies to $B + C = \pi / 2$.In $\triangle ABC$,$A + B + C = \pi$.Substituting $B + C = \pi / 2$,we get $A + \pi / 2 = \pi$,so $A = \pi / 2$.

Class 11 Mathematics Trigonometrical Equations Mix Examples-Trigonometrical Equations and Inequations, Properties of Triangles, Height and Distance

In a triangle $ABC$,$\angle B < \angle C$ and the values of $B$ and $C$ satisfy the equation $2 \tan x - k (1 + \tan^2 x) = 0$,where $0 < k < 1$. Then the measure of angle $A$ is:

A
$\pi / 3$
B
$2\pi / 3$
C
$\pi / 2$
D
$3\pi / 4$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Mathematics Questions

Chapter

Trigonometrical Equations

Subtopic

Mix Examples-Trigonometrical Equations and Inequations, Properties of Triangles, Height and Distance

In any triangle $ABC$,find the value of $\cos ^2 \frac{A}{2}+\cos ^2 \frac{B}{2}+\cos ^2 \frac{C}{2}$.

EasyTS EAMCET 2020

View Solution

For a $\Delta ABC$,if $a \cos^2 \frac{C}{2} + c \cos^2 \frac{A}{2} = \frac{3b}{2}$,then the sides $a, b, c$ are in:

Difficult

View Solution

If the perpendicular $AD$ divides the base of the triangle $ABC$ such that $BD, CD$ and $AD$ are in the ratio $2:3:6$,then angle $A$ is equal to

Easy

View Solution

In a $\Delta ABC$,if $\frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c}$ and the side $a = 2$,then the area of the triangle is:

MediumIIT 1993

View Solution

The product of the arithmetic mean of the lengths of the sides of a triangle and the harmonic mean of the lengths of the altitudes of the triangle is equal to: [where $\Delta$ is the area of the triangle $ABC$]

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

In a triangle $ABC$,$\angle B < \angle C$ and the values of $B$ and $C$ satisfy the equation $2 \tan x - k (1 + \tan^2 x) = 0$,where $0 < k < 1$. Then the measure of angle $A$ is:

Similar Questions

In any triangle $ABC$,find the value of $\cos ^2 \frac{A}{2}+\cos ^2 \frac{B}{2}+\cos ^2 \frac{C}{2}$.

For a $\Delta ABC$,if $a \cos^2 \frac{C}{2} + c \cos^2 \frac{A}{2} = \frac{3b}{2}$,then the sides $a, b, c$ are in:

If the perpendicular $AD$ divides the base of the triangle $ABC$ such that $BD, CD$ and $AD$ are in the ratio $2:3:6$,then angle $A$ is equal to

In a $\Delta ABC$,if $\frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c}$ and the side $a = 2$,then the area of the triangle is:

The product of the arithmetic mean of the lengths of the sides of a triangle and the harmonic mean of the lengths of the altitudes of the triangle is equal to: [where $\Delta$ is the area of the triangle $ABC$]

Vedclass Test Series

Exam Paper Generator

Online Exam Module