Question

sec theta + tan theta = p , show that sin theta = p^2 - 1 ÷ p^2 +1

Answer 1

this is required answer

Answer 2

Given :

If sec $\theta$ + tan $\theta$ = p, then show that sin $\theta$ = $\large\dfrac {p² - 1} {p² + 1}$

Solution :

$\normalsize\sf\ Given \: sec \: \theta \: + \: tan \: \theta \: = \: p \: \rule {25mm}{1pt} \: {\boxed{\sf{\red{1}}}}$

$\sf\ We \: know \: that \: {sec}^{2} \: \theta \: - \: {tan}^{2} \: \theta \: = \: 1$

$\sf\ ( \: sec \: \theta \: + \: tan \: \theta \: ) \: ( \: sec \: \theta \: - \: tan \: \theta \: ) \: = \: 1$

$\sf\ ⇒ \: p \: ( \: sec \: \theta \: - \: tan \: \theta \: ) \: = \: 1$

$\sf\ ⇒ \: sec \: \theta \: - \: tan \: \theta \: \: = \: {\large{\dfrac{1}{p}}} \: \rule {25mm}{1pt} \: {\boxed{\mathcal{\red{2}}}}$

Now, Add Eq 1 + 2

$\sf\ sec \: \theta \: \cancel{+ \: tan \: \theta} \: = \: p \\ \dfrac{\sf\ sec \: \theta \: \cancel{ - \: tan \: \theta} \: = \: {{\sf{ \dfrac{1}{p}}}}}{\sf\ \: 2 \: sec \: \theta \: = \: p \: + {{\sf{ \dfrac{1}{p}}}}}$

$\sf\ ⇒ \: 2 \: sec \: \theta \: = \: {{\sf{ \dfrac{ {p}^{2} \: + \: 1}{p}}}}$

$\sf\ ⇒ \: sec \: \theta \: = \: {{\sf{ \dfrac{ {p}^{2} \: + \: 1}{2p}}}} \: \rule {25mm}{1pt} \: {\boxed{\red{\sf{3}}}}$

$\sf\ sec \: \theta \: + \: tan \: \theta \: = \: p \\ \dfrac{\sf\ sec \: \theta \: - \: tan \: \theta \: = \: \dfrac{1}{p}}{\sf\ ⇒ \: 2 \: tan \: \theta \: \: = \: p \: - \: \dfrac{1}{p}}$

$\sf\ ⇒ \: 2 \: tan \: \theta \: = \: {{\dfrac{ {p}^{2} \: - \: 1}{p}}}$

$\sf\ ⇒ \: tan \: \theta \: = \: {{ \dfrac{ {p}^{2} \: - \: 1}{2p}}} \: \rule {25mm}{1pt} \: {\boxed{\sf{\red{4}}}}$

Now, divide equation 4 and 3 " $\sf\ ( \dfrac {4}{3} )$ "

$\sf\ \dfrac{4}{3} \: = \: \dfrac{tan \: \theta}{sec \: \theta} \: = \: {{\dfrac{ \dfrac{ {p}^{2} \: - \: 1}{\cancel{2p}}}{ \dfrac{ {p}^{2} \: + \: 1}{\cancel{2p}}}}}$

$\sf\ ⇒ \: \dfrac{ \dfrac{sin \: \theta}{cos \: \theta} }{cos \: \theta} \: = \: \dfrac{ {p}^{2} \: - \: 1}{ {p}^{2} \: + \: 1}$

${\large{\sf{\red{⇒}}}} \: \: \: {\large{\boxed{\sf{\red{sin \: \theta \: = \: {p}^{2} \: - \: 1}}}}}$

$\rule{100mm}{1pt}$

Know More :

Identidites :

$\sf\ \: \: sin^{2} \: \theta \: + \: cos {}^{2} \: \theta \: = \: 1$

$\sf\ \: \: sec {}^{2} \: \theta \: - \: tan {}^{2} \: \theta \: = \: 1$

$\sf\ \: \: {cosec}^{2} \: \theta \: - \: {cot}^{2} \: \theta \: = \: 1$

$\tiny$

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