Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

# If 3 Tan ( X − 15 ∘ ) = Tan ( X + 15 ∘ ) 0 < X < 90 ∘ , Find θ. - Mathematics

Sum

If $3\tan\left( x - 15^\circ \right) = \tan\left( x + 15^\circ \right)$ $0 < x < 90^\circ$, find θ.

#### Solution

Given: $3\tan\left( x - 15^\circ \right) = \tan\left( x + 15^\circ \right)$
$\Rightarrow \frac{\tan\left( x + 15^\circ \right)}{\tan\left( x - 15^\circ \right)} = 3$
Applying componendo and dividendo, we have
$\frac{\tan\left( x + 15^\circ \right) + \tan\left( x - 15^\circ \right)}{\tan\left( x + 15^\circ \right) - \tan\left( x - 15^\circ \right)} = \frac{3 + 1}{3 - 1}$
$\Rightarrow \frac{\frac{\sin\left( x + 15^\circ \right)}{\cos\left( x + 15^\circ \right)} + \frac{\sin\left( x - 15^\circ \right)}{\cos\left( x - 15^\circ \right)}}{\frac{\sin\left( x + 15^\circ \right)}{\cos\left( x + 15^\circ \right)} - \frac{\sin\left( x - 15^\circ \right)}{\cos\left( x - 15^\circ \right)}} = \frac{4}{2}$
$\Rightarrow \frac{\sin\left( x + 15^\circ \right)\cos\left( x - 15^\circ \right) + \cos\left( x + 15^\circ \right)\sin\left( x - 15^\circ \right)}{\sin\left( x + 15^\circ \right)\cos\left( x - 15^\circ \right) - \cos\left( x + 15^\circ \right)\sin\left( x - 15^\circ \right)} = 2$
$\Rightarrow \frac{\sin\left( x + 15^\circ + x - 15^\circ \right)}{\sin\left( x + 15^\circ- x + 15^\circ \right)} = 2$

$\Rightarrow \frac{\sin2x}{\sin30^\circ} = 2$

$\Rightarrow \sin2x = 2 \times \frac{1}{2} = 1 \left( \sin30^\circ = \frac{1}{2} \right)$

$\Rightarrow \sin2x = \sin90^\circ$

$\Rightarrow 2x = 90^\circ \left( 0 < x < 90^\circ \right)$

$\Rightarrow x = 45^\circ$

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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 11 Trigonometric equations
Q 11 | Page 26