Question

Prove that : 1-cosA/1+cosA =(cosecA - cotA ) square ?

•Answer with full explanation!
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Answer 1

Corrected Question :

• ${\sf{Prove~that~:~\dfrac{1~-~\cos A}{1~+~\cos A}~=~(\cot A~-~\cosec A)^2~\large{?}}}$

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Given : ${\sf{\dfrac{1~- ~\cos A}{1~+~ \cos A}~=~(\cot A~-~\cosec A)^2}}$

To Prove : ${\sf{\dfrac{1~-~\cos A}{1~+~\cos A}~=~\cos A~-~\cosec A)^2}}$

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${\sf{\dfrac{1~-~\cos A}{1~+~\cos A}~=~\cos A~-~\cosec A)^2}}$

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Here,

$~~~~~~~~~~{\sf:\implies{L.H.S~=~\dfrac{1~-~\cos A}{1~+~\cos A}}}$

$~~~~~~~~~~{\sf:\implies{R.H.S~=~(\cos A~-~\cosec A}}$

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$\underline{\frak{Now~ By~ Solving~ the ~L.H.S~:}}$

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$~~~~~~~~~~{\sf:\implies{L.H.S~=~\dfrac{1~-~\cos A}{1~+~\cos A}}}$

$~~~~~~~~~~{\sf:\implies{\dfrac{1~-~\cos A}{1~+~\cos A}}}$

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$\underline{\sf{Multiplying~both~and~ numerator ~and~ denominator ~by~\bf{(1~-~cos A)}}}$

$~~~~~~~~~~{\sf:\implies{\dfrac{1~-~\cos A}{1~+~\cos A}~×~\dfrac{1~-~\cos A}{1~-~\cos A}}}$

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$\underline{\frak{As ~we ~know~ that~:}}$

• $\boxed{\sf\pink{(a~+~b)(a~-~b)~=~(a~-~b)^2}}$★

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$~~~~~~~~~~{\sf:\implies{\dfrac{1~-~\cos A}{1~+~\cos A}~×~\dfrac{1~-~\cos A}{1~-~\cos A}}}$

$~~~~~~~~~~{\sf:\implies{\dfrac{(1~-~\cos A)^{2}}{1~-~\cos^{2} A}}}$

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$\underline{\frak{As ~we ~know ~that~:}}$

• $\boxed{\sf\pink{\sin^{2}\theta~=~1~-~\cos^{2} A}}$★

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$~~~~~~~~~~{\sf:\implies{\dfrac{(1~-~cos A)^2}{sin^{2} A}}}$

$~~~~~~~~~~{\sf:\implies{\bigg(\dfrac{1~-~\cos A}{sin A}\bigg)^2}}$

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Or,

$~~~~~~~~~~{\sf:\implies{\bigg(\dfrac{1}{\sin A}~-~\dfrac{\cos A}{\sin A}\bigg)^2}}$

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$\underline{\frak{As~ we~ know~ that~:}}$

• $\boxed{\sf\pink{\dfrac{1}{\sin A}~=~cosec A}}$★

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$~~~~~~~~~~{\sf:\implies{\bigg(cosec A~-~\dfrac{\cos A}{\sin A}\bigg)^2}}$

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$\underline{\frak{As ~we ~know~ that~:}}$

• $\boxed{\sf\pink{\dfrac{\cos A}{\sin A}~=~\cot A}}$★

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$~~~~~~~~~~{\sf:\implies{\bigg(\cosec A~-~cot\bigg)^2}}$

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Or,

$~~~~~~~~~~{\sf:\implies{L.H.S~=~\bigg(\cot A~-~\cosec A\bigg)^2}}$

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Therefore,

$~~~~~~~~~~{\sf:\implies{L.H.S~=~\bigg(\cot A~-~cosec\bigg)^2}}$

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$~~~~~~~~~~{\sf:\implies{L.H.S~=~\bigg(\cot A~-~\cosec \bigg)^2}}$

$~~~~~~~~~~{\sf:\implies{R.H.S~=~(\cot A~-~\cosec A)^2}}$

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Hence,

$~~~~~~~~~~~~~~~~~~$$\boxed{\sf{L.H.S~=~R.H.S}}$

$~~~~$$\qquad\quad\therefore\underline{\textsf{\textbf{Hence Proved!}}}$