Question

prove that -: secA -1 / secA +1=sin^2A/(1+cos)^2​

Answer 1

Solution:-

$\implies\sf\ \dfrac{secA-1}{secA+1}= \dfrac{sin^2A}{(1+cosA)^2}\\ \\ \\ \ \ taking\ LHS\\ \\ \\ :\implies\sf\ Multiplying\ both\ Numerator\ and\ denominator\ by\ secA+1\\ \\ \\ :\implies\sf\ \dfrac{secA-1}{secA+1}\times\dfrac{secA+1}{secA+1}\\ \\ \\ :\implies\sf\ \dfrac{sec^2A-1}{(secA+1)^2}\\ \\ \\ :\implies\sf\ \ \dfrac{tan^2A}{(1/_{cosA}+1)^2}\\ \\ \\ :\implies\sf\ \dfrac{tan^2A}{^{(1+cosA)^2}/_{cos^2A}}\\ \\ \\ :\implies\sf\ \dfrac{^{sin^2A}/_{cos^2A}}{^{(1+cosA)^2}/_{cos^2A}}\\ \\ \\ :\implies\sf\ \dfrac{sin^2A}{\cancel{cos^2A}}\times\ \dfrac{\cancel{cos^2A}}{(1+cosA)^2}\\ \\ \\ :\implies\pink{\sf\ \ \dfrac{sin^2A}{(1+cosA)^2}}\ \ \ \ \ \sf\ Hence\ Proved !!$

Answer 2

$\; \; \; \dag{\large{\bold{\red{\sf{Required \; solution}}}}}$

$\begin{gathered}\rightarrow\bf\ \dfrac{secA - 1}{secA + 1} = \dfrac{sin^{2}A}{(1 + cosA)^{2}}\\ \\ \bf Now \ let \ us \ take \ Left \ Hand \ Side \\ \\ \bf\ Now \ let's \ multiply \ both \ the \ numerator \ and \ denominator \ by \ secA + 1\\ \\ \rightarrow \bf\ \dfrac{secA - 1}{secA + 1}\times\dfrac{secA + 1}{secA + 1}\\ \\ \rightarrow \bf\ \dfrac{sec^{2}A - 1}{(secA + 1)^{2}}\\ \\ \rightarrow \sf\ \ \dfrac{tan^{2}A}{(1/_{cosA}+1)^2}\\ \\ \rightarrow \bf\ \dfrac{tan^2A}{^{(1+cosA)^2}/_{cos^2A}}\\ \\ \rightarrow \bf \ \dfrac{^{sin^2A}/_{cos^2A}}{^{(1+cosA)^2}/_{cos^2A}}\\ \\ \rightarrow \bf\ \dfrac{sin^2A}{cos^2A}\times\ \dfrac{cos^2A}{(1+cosA)^2}\\ \\ \rightarrow {\bf\ \ \dfrac{sin^2A}{(1+cosA)^2}} \end{gathered}$

Hence proved !..

${\large{\bold{\red{\sf{Additional \; information}}}}}\; \; \; \dag$

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

Some important trigonometry identity -

• cosec²θ - cot²θ = 1

• cosec²θ = 1 + cot²θ

• 1 + cot²θ = cosec²θ

• sin²θ + cos²θ = 1

• sin²θ = 1 - cos²θ

• cos²θ = 1 - sin²θ

• sec²θ = 1 + tan²θ

• sec²θ - tan²θ = 1