Question

Sin A/1 +Cot A - Cos A/1+Tan A =Sina A - Cos A

Answer 1

Question :

Prove that,

$\sf{\dfrac{sinA}{1+cotA}-\dfrac{cosA}{1+tanA}=sinA-cosA}$

Solution :

Taking L.H.S,

$\sf{\dfrac{sinA}{1+cotA}-\dfrac{cosA}{1+tanA}}$

Write $\sf{cotA=\frac{cosA}{sinA}}$ and $\sf{tanA=\frac{sinA}{cosA}}$

$\implies\sf{\dfrac{sinA}{1+\dfrac{cosA}{sinA}}-\dfrac{cosA}{1+\dfrac{sinA}{cosA}}}$

$\implies\sf{\dfrac{sinA}{\dfrac{sinA+cosA}{sinA}}-\dfrac{cosA}{\dfrac{cosA+sinA}{cosA}}}$

$\implies\sf{sinA\times\dfrac{sinA}{sinA+cosA}-cosA\times\dfrac{cosA}{cosA+sinA}}$

$\implies\sf{\dfrac{sin^2A}{sinA+cosA}-\dfrac{cos^2A}{sinA+cosA}}$

$\implies\sf{\dfrac{sin^2A-cos^2A}{sinA+cosA}}$

Use the identity : a²-b² = ( a+b)(a-b)

$\implies\sf{\dfrac{(sinA+cosA)(sinA-cosA)}{sinA+cosA}}$

$\implies\sf{sinA-cosA}$

L.H.S = R.H.S [ Proved ]

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Some identities related to trigonometry :

• sin²A + cos²A = 1

• 1 + tan²A = sec²A

• 1 + cot²A = cosec²A

• cos²A - sin²A = cos2A

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Answer 2

Question :–

▪︎Prove that : $\\ \dfrac{ \sin(A) }{1 + \cot(A) } - \dfrac{ \cos(A) }{1 + \tan(A) } =\sin(A) - \cos(A)$

ANSWER :–

Taking L.H.S. –

$\\ = \dfrac{ \sin(A) }{1 + \cot(A) } - \dfrac{ \cos(A) }{1 + \tan(A) }$

• We know that –

$\\ \implies \tan(A) = \dfrac{ \sin(A) }{ \cos(A) } \: \: and \: \: \cot(A) = \dfrac{ \cos(A) }{ \sin(A) }$

• So , that –

$\\ = \dfrac{ \sin(A) }{1 +[ \dfrac{ \cos(A) }{ \sin(A) }]} - \dfrac{ \cos(A) }{1 + [ \dfrac{ \sin(A) }{ \cos(A) } ]}$

$\\ = \dfrac{ \sin(A) }{ \dfrac{ \sin(A) + \cos(A) }{ \sin(A) }} - \dfrac{ \cos(A) }{ \dfrac{\cos(A) + \sin(A) }{ \cos(A) } }$

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

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

• We know that –

$\\ \implies { \bold{ {a}^{2} - {b}^{2} = (a + b)(a - b) }}$

• So , that –

$\\ = \dfrac{[\sin (A) - \cos (A)] [\sin(A) + \cos(A)]}{[\sin(A) + \cos(A)]}$

$\\ = \dfrac{[\sin (A) - \cos (A)] {\cancel{ [\sin(A) + \cos(A)]}}}{ \cancel{[\sin(A) + \cos(A)]}}$

$\\ = \sin (A) - \cos (A)$

= R.H.S. (Hence proved)