Question

if tan theta-4/tan theta = 3 then sin2 theta is​....please answer this with detailed solution

Answer 1

Given,

$\[\tan \theta - \frac{4}{{\tan \theta }} = 3\]$

Solution,

Calculate the value of $\[\tan \theta \]$

$\[\tan \theta - \frac{4}{{\tan \theta }} = 3\]$

$\[\begin{array}{l} \Rightarrow {\tan ^2}\theta - 4 = 3\tan \theta \\ \Rightarrow {\tan ^2}\theta - 3\tan \theta - 4 = 0\\ \Rightarrow {\tan ^2}\theta - 4\tan \theta + \tan \theta - 4 = 0\\ \Rightarrow \tan \theta \left( {\tan \theta - 4} \right) + 1\left( {\tan \theta - 4} \right) = 0\\ \Rightarrow \left( {\tan \theta + 1} \right)\left( {\tan \theta - 4} \right) = 0\\ \Rightarrow \tan \theta = - 1,4\end{array}\]$

Now calculate the value of $\[sin2\theta \]$.

When $\[\tan \theta = - 1\]$,

$\[\begin{array}{l} = sin2\theta \\ = \frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}\\ = \frac{{2 \times \left( { - 1} \right)}}{{1 + {{\left( { - 1} \right)}^2}}}\\ = - 2\end{array}\]$

When$\[\tan \theta = 4\]$,

$\[\begin{array}{l} = sin2\theta \\ = \frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}\\ = \frac{{2 \times 4}}{{1 + {4^2}}}\\ = \frac{8}{{17}}\end{array}\]$

Hence the value of $\[sin2\theta \]$ will be $\[ - 2,\frac{8}{{17}}\]$.

Answer 2

Given:

$\[\tan \theta - \frac{4}{{\tan \theta }} = 3\]$

Solution:

Simplify the given equation.

$\[\begin{array}{l}\tan \theta - \frac{4}{{\tan \theta }} = 3\\ \Rightarrow {\tan ^2}\theta - 4 = 3\tan \theta \\ \Rightarrow {\tan ^2}\theta - 3\tan \theta - 4 = 0\end{array}\]$------(1)

Assume that, $\[x = \tan \theta \]$.

Substitute in equation (1).

$\[\begin{array}{l} \Rightarrow {x^2} - 3x - 4 = 0\\ \Rightarrow {x^2} - 4x + x - 4 = 0\\ \Rightarrow x\left( {x - 4} \right) + 1\left( {x - 4} \right) = 0\\ \Rightarrow x = - 1,4\end{array}\]$

Therefore,

$\[ \Rightarrow \tan \theta = - 1,4\]$

Know that, $\[\tan \theta = \frac{{\sin \theta }}{{\cos \theta }}\]$.

Take reciprocal and square both the sides.

$\[\begin{array}{l} \Rightarrow \frac{1}{{\tan \theta }} = \frac{{\cos \theta }}{{\sin \theta }}\\ \Rightarrow {\left( {\frac{1}{{\tan \theta }}} \right)^2} = \frac{{{{\cos }^2}\theta }}{{{{\sin }^2}\theta }}\end{array}\]$

Add 1 both sides on the above equation,

$\[\begin{array}{l} \Rightarrow {\left( {\frac{1}{{\tan \theta }}} \right)^2} + 1 = \frac{{{{\cos }^2}\theta }}{{{{\sin }^2}\theta }} + 1\\ \Rightarrow {\left( {\frac{1}{{\tan \theta }}} \right)^2} + 1 = \frac{{{{\cos }^2}\theta + {{\sin }^2}\theta }}{{{{\sin }^2}\theta }}\\ \Rightarrow {\left( {\frac{1}{{\tan \theta }}} \right)^2} + 1 = \frac{1}{{{{\sin }^2}\theta }}\end{array}\]$

$\[ \Rightarrow {\sin ^2}\theta = \frac{1}{{{{\left( {\frac{1}{{\tan \theta }}} \right)}^2} + 1}}\]$

Take$\[\tan \theta = - 1\]$.

$\[ \Rightarrow {\sin ^2}\theta = \frac{1}{{{{\left( {\frac{1}{{ - 1}}} \right)}^2} + 1}}\]$

$\[ \Rightarrow {\sin ^2}\theta = \frac{1}{{1 + 1}}\]$

$\[ \Rightarrow {\sin ^2}\theta = \frac{1}{2}\]$

Take $\[\tan \theta = 4\]$.

$\[ \Rightarrow {\sin ^2}\theta = \frac{1}{{{{\left( {\frac{1}{4}} \right)}^2} + 1}}\]$

$\[ \Rightarrow {\sin ^2}\theta = \frac{1}{{\frac{1}{{16}} + 1}}\]$

$\[ \Rightarrow {\sin ^2}\theta = \frac{1}{{\frac{{1 + 16}}{{16}}}}\]$

$\[ \Rightarrow {\sin ^2}\theta = \frac{{16}}{{17}}\]$

Hence, the values of $\[{\sin ^2}\theta \]$are $\[\frac{1}{2},\frac{{16}}{{17}}\]$.