Question

qυєsтıση :-
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Prove the identity :
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$\bf{1 - \dfrac{sin^2\: x}{1 + cot\: x} - \dfrac{cos^2\: x}{1 + tan\: x} = sin\: x .cos\: x}$
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Answer 1

Topic :-

Trigonometry

To Prove :-

$1-\dfrac{\sin^2x}{1+\cot x}-\dfrac{\cos^2x}{1+\tan x}=\sin x .\cos x$

Solution :-

Solving LHS,

$1-\dfrac{\sin^2x}{1+\cot x}-\dfrac{\cos^2x}{1+\tan x}$

$1-\dfrac{\sin^2x}{1+\dfrac{\cos x}{\sin x}}-\dfrac{\cos^2x}{1+\dfrac{\sin x}{\cos x}}$

$1-\dfrac{\sin^3x}{\sin x+\cos x}-\dfrac{\cos^3x}{\sin x+\cos x}$

$\dfrac{(\sin x +\cos x)-(\sin^3 x +\cos^3 x)}{\sin x +\cos x}$

$\mathtt {a^3+b^3=(a+b)(a^2-ab+b^2)}$

$\dfrac{(\sin x +\cos x)-(\sin x +\cos x)(\sin^2x-\sin x \cos x+ \cos^2x)}{\sin x +\cos x}$

$\mathtt {\sin^2x+\cos^2x=1}$

$\dfrac{(\sin x +\cos x)-(\sin x +\cos x)(1-\sin x \cos x)}{\sin x +\cos x}$

$\dfrac{(\sin x+ \cos x)(1-(1-\sin x. \cos x))}{(\sin x+\cos x)}$

$\mathtt {1-(1-\sin x. \cos x)}$

$\mathtt {1-1+\sin x.\cos x}$

$\mathtt {\sin x. \cos x}$

RHS,

$\mathtt {\sin x. \cos x}$

We can observe that LHS = RHS,

Hence, Proved !

Additional Formulae :-

$\mathtt{1+\tan^2x=\sec^2x}$

$\mathtt{1+\cot^2x=cosec^2x}$

$\mathtt {cos^4x-sin^4x=cos(2x)}$

$\mathtt {1+cos2x=2cos^2x}$

$\mathtt {1-cos2x=2sin^2x}$

Answer 2

$\huge\mathrm{Notes}$

⭐ Some major formulae of trigonometry are ➡

cos²x + sin²x = 1

1 + tan²x = sec²x

1 + cot²x = cosec²x

⭐ Some major formula applied in the answer are ➡

a³ + b³ = ( a + b )( a² - ab + b² )

⭐ The answer should be written in step by step order ➡

• First you must write To proof :- and then below that you should write the proof
• Secondly you should write Proof - and then write L.H.S ( left hand side ) and then write the left hand equation and then start proofing it
• Write the reason and formulae wherever required
• After you successfully do the proof write = R.H.S ( Right hand side ) , the one you have to proof
• And finally complete the answer by writing ➡

L.H.S = R.H.S ( Proved )

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