Math, asked by Anonymous, 5 months ago

cosec²θ / cosec θ -1 - cosec²θ / cosecθ + 1 = 2 sec²θ
prove it plsssssssssssssss​

Answers

Answered by mathdude500
0

Question :-

\bf \:Prove \: that \: \dfrac{ {cosec}^{2}θ }{cosecθ - 1}  - \dfrac{ {cosec}^{2}θ }{cosecθ + 1}  =  {2sec}^{2} θ

Solution :-

\bf \:Consider \: \dfrac{ {cosec}^{2}θ }{cosecθ - 1}  - \dfrac{ {cosec}^{2}θ }{cosecθ + 1}

\bf\implies \: {cosec}^{2} θ(\dfrac{ 1 }{cosecθ - 1}  - \dfrac{ 1 }{cosecθ + 1})

\bf\implies \: {cosec}^{2} θ(\dfrac{cosecθ + 1 - cosecθ + 1}{(cosecθ + 1)(cosecθ - 1)} )

\bf\implies \:\dfrac{ {2cosec}^{2} θ}{ {cosec}^{2}θ - 1 }

\bf\implies \:\dfrac{2 {cosec}^{2}θ }{ {cot}^{2}θ }

\bf\implies \:2 \times \dfrac{1}{ {sin}^{2}θ }  \times \dfrac{ {sin}^{2} θ}{ {cos}^{2}θ }

\bf\implies \:\dfrac{2}{ {cos}^{2} θ}

\bf\implies \: {2sec}^{2} θ

\large{\boxed{\boxed{\bf{Hence, Proved}}}}

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❥︎Additional Information:-

Relationship between sides and T ratios

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

❥︎Reciprocal Identities

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

❥︎Co-function Identities

sin (90°−x) = cos x

cos (90°−x) = sin x

tan (90°−x) = cot x

cot (90°−x) = tan x

sec (90°−x) = cosec x

cosec (90°−x) = sec x

❥︎Fundamental Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1

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

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