Question

Prove it (1+cosec A) / cosec A = sin^2A/(1-sin A)


Note :- First find the value of A then prove that
(1+cosec A) / cosec A = sin^2A/(1-sin A) (Putting the value of A)

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Answer 1

ㅤ$\;\;\underline{\textbf{\textsf{ Given :-}}}$

$\sf\dashrightarrow \dfrac{1 + cosecA}{cosecA} = \dfrac{sin^2A}{1 - sinA}$

ㅤ$\;\;\underline{\textbf{\textsf{ To Prove :-}}}$

• L.H.S = R.H.S

ㅤ$\;\;\underline{\textbf{\textsf{ Proof :-}}}$

$\bf\underline{\green{\:\:\:\:\:\:\:\:A.T.Q:-\:\:\:\:\:\:\:}}$

We need to find the value of A.

$\sf \dashrightarrow \dfrac{1 + cosecA}{cosecA} = \dfrac{sin^2A}{1 - sinA}$

$\sf \dashrightarrow \dfrac{1}{cosecA} + \dfrac{cosecA}{cosecA} = \dfrac{sin^2A}{1 - sinA}$

$\sf \dashrightarrow sinA + 1 = \dfrac{sin^2A}{1 - sinA}$

$\sf \dashrightarrow (sinA + 1)(1 - sinA) = sin^2A$

$\sf \dashrightarrow (1)^2 - (sinA)^2 = sin^2A$

$\sf \dashrightarrow 1 - sin^2A = sin^2A$

$\sf \dashrightarrow 1 = sin^2A + sin^2A$

$\sf \dashrightarrow 1 = 2sin^2A$

$\sf \dashrightarrow 2sin^2A = 1$

$\sf \dashrightarrow sin^2A = \dfrac{1}{2}$

$\sf\dashrightarrow sinA = \sqrt{\dfrac{1}{2}}$

$\sf \dashrightarrow sinA = \dfrac{1}{\sqrt{2}}$

$\sf \dashrightarrow A = sin^{-1} \ \dfrac{1}{\sqrt{2}}$

$\sf \dashrightarrow A = 45 ^\circ$

ㅤ$\;\;\underline{\textbf{\textsf{ Hence-}}}$

$\underline{\textsf{ Value of A is \textbf{45°}}}.$

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Again, we need to prove L.H.S and R.H.S

$\sf\dashrightarrow \dfrac{1 + cosecA}{cosecA} = \dfrac{sin^2A}{1 - sinA}$

L.H.S

$\sf \dashrightarrow \dfrac{1 + cosec45^\circ}{cosec45^\circ}$

$\sf \dashrightarrow \dfrac{1 + \sqrt{2}}{\sqrt{2}}$

$\sf\dashrightarrow \dfrac{1 + \sqrt{2}}{\sqrt{2}} \times \dfrac{\sqrt{2}}{\sqrt{2}}$

$\sf \dashrightarrow \dfrac{\sqrt{2} + (\sqrt{2})^2}{(\sqrt{2})^2}$

$\sf \dashrightarrow \dfrac{\sqrt{2} + 2}{2}$

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R.H.S

$\sf\dashrightarrow \dfrac{sin^2A}{1 - sinA}$

$\sf \dashrightarrow \dfrac{sin^245^\circ}{1 - sin45^\circ}$

$\sf \dashrightarrow \dfrac{\bigg(\dfrac{1}{\sqrt{2}}\bigg)^2}{1 - \bigg(\dfrac{1}{\sqrt{2}}\bigg)}$

$\sf \dashrightarrow \dfrac{\dfrac{1}{2}}{\bigg(\dfrac{\sqrt{2} - 1}{\sqrt{2}}\bigg)}$

$\sf \dashrightarrow \dfrac{1}{2} \times \dfrac{\sqrt{2}}{\sqrt{2} - 1}$

$\sf \dashrightarrow \dfrac{\sqrt{2}}{2(\sqrt{2} - 1)}$

$\sf\dashrightarrow \dfrac{\sqrt{2}}{2\sqrt{2} - 2}$

$\sf \dashrightarrow \dfrac{\sqrt{2}}{2\sqrt{2} - 2} \times \dfrac{2\sqrt{2} + 2}{2\sqrt{2} + 2}$

$\sf \dashrightarrow \dfrac{\sqrt{2}(2\sqrt{2} + 2)}{(2\sqrt{2})^2 - (2)^2}$

$\sf \dashrightarrow \dfrac{ \sqrt{2}(2\sqrt{2}) + \sqrt{2}(2)}{(4 \times 2) - 4}$

$\sf \dashrightarrow \dfrac{4 + 2\sqrt{2}}{8 - 4}$

$\sf\dashrightarrow \dfrac{2(2 + \sqrt{2})}{4}$

$\sf\dashrightarrow \dfrac{2 + \sqrt{2}}{2}$

ㅤ$\;\;\underline{\textbf{\textsf{Hence -}}}$

$\leadsto \ {\boxed{\tt{LHS = RHS}}}$

$\therefore{ \underline{\bf{(Proved)}}}$

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$\;\;\underline{\textbf{\textsf{ Know More :-}}}$

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$\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3} }{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }& 1 & \sqrt{3} & \rm Not \: De fined \\ \\ \rm cosec A & \rm Not \: De fined & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm Not \: De fined \\ \\ \rm cot A & \rm Not \: De fined & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0 \end{array}$

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Answer 2

Answer:

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