Question

prove that tan 45 - theta is equal to cos theta minus sin theta whole upon cos theta + sin theta​

Answer 1

QUESTION:

Prove that :

$\sf Tan \ (45^{\circ} - \theta)=\dfrac{Cos\ \theta-Sin\ \theta}{Cos\ \theta+Sin\ \theta}$

GIVEN:

$\sf Tan \ (45^{\circ} - \theta)=\dfrac{Cos\ \theta-Sin\ \theta}{Cos\ \theta+Sin\ \theta}$

TO PROVE:

$\sf Tan \ (45^{\circ} - \theta)=\dfrac{Cos\ \theta-Sin\ \theta}{Cos\ \theta+Sin\ \theta}$

EXPLANATION:

$\boxed{ \boldsymbol{ \gray{Tan \ (x - y)=\dfrac{Tan \ x - Tan \ y }{1 + Tan \ x \ Tan \ y}}}}$

$\sf Tan \ (45^{\circ} - \theta)=\dfrac{Tan \ 45^{\circ} - Tan \ \theta}{1 + Tan \ 45^{\circ} \ Tan \ \theta}$

$\boxed{ \boldsymbol{ \large {\gray{Tan \ 45^{\circ} = 1}}}}$

$\sf Tan \ (45^{\circ} - \theta)=\dfrac{1- Tan \ \theta}{1 + Tan \ \theta}$

$\boxed{ \boldsymbol{ \large {\gray{Tan \ \theta= \dfrac{Sin \ \theta}{Cos \ \theta} }}}}$

$\sf Tan \ (45^{\circ} - \theta)=\dfrac{1- \dfrac{Sin \ \theta}{Cos \ \theta}}{1 + \dfrac{Sin \ \theta}{Cos \ \theta}}$

$\sf Tan \ (45^{\circ} - \theta)=\dfrac{\dfrac{Cos \ \theta - Sin \ \theta}{Cos \ \theta}}{ \dfrac{Cos \ \theta + Sin \ \theta}{Cos \ \theta}}$

$\sf Tan \ (45^{\circ} - \theta)=\dfrac{Cos \ \theta - Sin \ \theta}{Cos \ \theta + Sin \ \theta}$

HENCE PROVED.

Trigonometric Ratio Table:

$\Large{ \begin{tabular}{|c|c|c|c|c|c|} \cline{1-6} \theta & \sf 0^{\circ} & \sf 30^{\circ} & \sf 45^{\circ} & \sf 65^{\circ} & \sf 90^{\circ} \\ \cline{1-6} \sin & 0 & \dfrac{1}{2 } & \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} & 1 \\ \cline{1-6} \cos & 1 & \dfrac{ \sqrt{ 3 }}{2} } & \dfrac{1}{ \sqrt{2} } & \dfrac{ 1 }{ 2 } & 0 \\ \cline{1-6} \tan & 0 & \dfrac{1}{ \sqrt{3} } & 1 & \sqrt{3} & \infty \\ \cline{1-6} \cot & \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } &0 \\ \cline{1 - 6} \sec & 1 & \dfrac{2}{ \sqrt{3}} & \sqrt{2} & 2 & \infty \\ \cline{1-6} \csc & \infty & 2 & \sqrt{2 } & \dfrac{ 2 }{ \sqrt{ 3 } } & 1 \\ \cline{1 - 6}\end{tabular}}$