Question

tan theta = 3/4 , find the value of cos theta + sin theta/cos theta- sin theta ?​

Answer 1

GIVEN :–

• tan(θ) = ¾

TO FIND :–

$\\\implies\bf Value\:\:of\:\:\dfrac{ \cos( \theta) + \sin( \theta) }{\cos( \theta) - \sin( \theta)} =?$

SOLUTION :–

$\\\implies\bf \tan( \theta) = \dfrac{3}{4}$

• We should write this as –

$\\\implies\bf \dfrac{\sin( \theta)}{\cos( \theta)}= \dfrac{3}{4}$

• We know that –

$\\\longrightarrow \red{\bf \sin^{2}( \theta) + \cos^{2} ( \theta) = 1}$

• So that –

$\\\implies\bf \dfrac{\sin( \theta)}{ \sqrt{1 - \sin^{2} ( \theta)}}= \dfrac{3}{4}$

• Square on both side –

$\\\implies\bf \left[\dfrac{\sin( \theta)}{ \sqrt{1 - \sin^{2} ( \theta)}} \right]^{2} = \left[\dfrac{3}{4} \right]^{2}$

$\\\implies\bf \dfrac{\sin^{2} ( \theta)}{1 - \sin^{2} ( \theta)}= \dfrac{9}{16}$

$\\\implies\bf16\sin^{2} ( \theta)= 9 \{1 - \sin^{2} ( \theta) \}$

$\\\implies\bf16\sin^{2} ( \theta)= 9 - 9\sin^{2} ( \theta)$

$\\\implies\bf16\sin^{2} ( \theta) + 9\sin^{2} ( \theta)= 9$

$\\\implies\bf25\sin^{2} ( \theta)= 9$

$\\\implies\bf\sin^{2} ( \theta)= \dfrac{9}{25}$

$\\\implies \large \red{ \boxed{\bf\sin ( \theta)= \dfrac{3}{5}}}$

• And –

$\\\implies\bf\cos( \theta)= \sqrt{1 - \sin^{2} ( \theta)}$

$\\\implies\bf\cos( \theta)= \sqrt{1 - \dfrac{9}{25} }$

$\\\implies\bf\cos( \theta)= \sqrt{ \dfrac{25 - 9}{25} }$

$\\\implies\bf\cos( \theta)= \sqrt{ \dfrac{16}{25} }$

$\\\implies \large \red{ \boxed{\bf\cos( \theta)= \dfrac{4}{5}}}$

• Now let's find –

$\\\implies\bf \dfrac{ \cos( \theta) + \sin( \theta) }{\cos( \theta) - \sin( \theta)} =\dfrac{\dfrac{4}{5}+ \dfrac{3}{5}}{\dfrac{4}{5}-\dfrac{3}{5}}$

$\\\implies\bf \dfrac{ \cos( \theta) + \sin( \theta) }{\cos( \theta) - \sin( \theta)} =\dfrac{\dfrac{4+3}{5}}{\dfrac{4 - 3}{5}}$

$\\\implies\bf \dfrac{ \cos( \theta) + \sin( \theta) }{\cos( \theta) - \sin( \theta)} =\dfrac{4+3}{4 - 3}$

$\\\implies\bf \dfrac{ \cos( \theta) + \sin( \theta) }{\cos( \theta) - \sin( \theta)} =\dfrac{7}{1}$

$\\\implies\bf \dfrac{ \cos( \theta) + \sin( \theta) }{\cos( \theta) - \sin( \theta)} =7$

• Hence , The value of $\bf \dfrac{ \cos( \theta) + \sin( \theta) }{\cos( \theta) - \sin( \theta)} =7$.

Answer 2

$\large\bf{\color{royalblue}{GIVEN :–}}$

• tan(θ) = ¾

$\large\bf{\color{purple}{TO FIND :–}}$

$\begin{gathered}\\\implies\bf Value\:\:of\:\:\dfrac{ \cos( \theta) + \sin( \theta) }{\cos( \theta) - \sin( \theta)} =?\\\end{gathered}$

$\large\bf{\pink{SOLUTION :–}}$

$\begin{gathered}\\\implies\bf \tan( \theta) = \dfrac{3}{4} \\\end{gathered}$

• We should write this as –

$\begin{gathered}\\\implies\bf \dfrac{\sin( \theta)}{\cos( \theta)}= \dfrac{3}{4} \\\end{gathered}$

• We know that –

$\begin{gathered}\\\longrightarrow \red{\bf \sin^{2}( \theta) + \cos^{2} ( \theta) = 1}\\\end{gathered}$

• So that –

$\begin{gathered}\\\implies\bf \dfrac{\sin( \theta)}{ \sqrt{1 - \sin^{2} ( \theta)}}= \dfrac{3}{4} \\\end{gathered}$

• Square on both side –

$\begin{gathered}\\\implies\bf [\dfrac{\sin( \theta)}{ \sqrt{1 - \sin^{2} ( \theta)}} ]^{2} = [\dfrac{3}{4} ]^{2} \\\end{gathered}$

$\begin{gathered}\\\implies\bf \dfrac{\sin^{2} ( \theta)}{1 - \sin^{2} ( \theta)}= \dfrac{9}{16}\\\end{gathered}$

$\begin{gathered}\\\implies\bf16\sin^{2} ( \theta)= 9 \{1 - \sin^{2} ( \theta) \}\\\end{gathered}$

$\begin{gathered}\\\implies\bf16\sin^{2} ( \theta)= 9 - 9\sin^{2} ( \theta)\\\end{gathered}$

$\begin{gathered}\\\implies\bf16\sin^{2} ( \theta) + 9\sin^{2} ( \theta)= 9\\\end{gathered}$

$\begin{gathered}\\\implies\bf25\sin^{2} ( \theta)= 9\\\end{gathered}$

$\begin{gathered}\\\implies\bf\sin^{2} ( \theta)= \dfrac{9}{25}\\\end{gathered}$

$\begin{gathered}\\\implies \large \red{ \boxed{\bf\sin ( \theta)= \dfrac{3}{5}}}\\\end{gathered}$

• And –

$\begin{gathered}\\\implies\bf\cos( \theta)= \sqrt{1 - \sin^{2} ( \theta)} \\\end{gathered}$

$\begin{gathered}\\\implies\bf\cos( \theta)= \sqrt{1 - \dfrac{9}{25} } \\\end{gathered}$

$\begin{gathered}\\\implies\bf\cos( \theta)= \sqrt{ \dfrac{25 - 9}{25} } \\\end{gathered}$