Question

prove that 1+sin theta/1-sin theta - 1-sin theta/1+sin theta= 4sec theta tan theta​

Answer 1

Answer:

Step-by-step explanation:

To Prove :-

$\dfrac{1+sin\theta}{1-sin\theta}-\dfrac{1-sin\theta}{1+sin\theta}=4sec\theta tan\theta$

Solution :-

Taking L.H.S :-

$=\dfrac{1+sin\theta}{1-sin\theta}-\dfrac{1-sin\theta}{1+sin\theta}$

$=\dfrac{(1+sin\theta)^2-(1-sin\theta)^2}{(1)^2-sin^2\theta}$

$=\dfrac{1+sin^2\theta+2sin\theta-(1+sin^2\theta-2sin\theta)}{1-sin^2\theta}$

$=\dfrac{1+sin^2\theta+2sin\theta-1-sin^2\theta+2sin\theta}{cos^2\theta}$

$=\dfrac{4sin\theta}{cos^2\theta}$

$= 4\times\dfrac{sin\theta}{cos\theta}\times\dfrac{1}{cos\theta}$

$=4\times tan\theta\times sec\theta$

$= 4sec\theta tan\theta$

= R.H.S

Hence Proved

Know More :-

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$