If A, B and C are the angles of a triangle such that A: B: C=2: 3: 4, then what is the value of the following?
$$ \frac{\tan^2 B+1}{\tan^2 B-1} $$
4
2
1
0
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Answer is Wrong!
This question was previously asked in
UPSC CBI DSP LDCE – 2023
The sum of angles in a triangle is 180°.
Let the common ratio factor be $x$. Then $A = 2x$, $B = 3x$, and $C = 4x$.
$2x + 3x + 4x = 180°$
$9x = 180°$
$x = \frac{180°}{9} = 20°$.
The angles are:
A = $2x = 2 \times 20° = 40°$
B = $3x = 3 \times 20° = 60°$
C = $4x = 4 \times 20° = 80°$
We need to find the value of the expression $\frac{\tan^2 B+1}{\tan^2 B-1}$.
Substitute $B = 60°$:
$\tan B = \tan 60° = \sqrt{3}$.
$\tan^2 B = (\sqrt{3})^2 = 3$.
Now substitute this value into the expression:
$\frac{\tan^2 B+1}{\tan^2 B-1} = \frac{3+1}{3-1} = \frac{4}{2} = 2$.
$\frac{\tan^2 B+1}{\tan^2 B-1} = \frac{\sec^2 B}{\frac{\sin^2 B}{\cos^2 B}-1} = \frac{\frac{1}{\cos^2 B}}{\frac{\sin^2 B – \cos^2 B}{\cos^2 B}} = \frac{1}{\sin^2 B – \cos^2 B}$.
Using the identity $\cos(2B) = \cos^2 B – \sin^2 B = -(\sin^2 B – \cos^2 B)$, the expression is $\frac{1}{-\cos(2B)}$.
For $B=60°$, $2B=120°$. $\cos(120°) = -\frac{1}{2}$.
The value is $\frac{1}{-(-\frac{1}{2})} = \frac{1}{\frac{1}{2}} = 2$.