Question

Prove that:-
$\frac{ \cos(a) }{1 - \tan(a) } + \frac{ \sin(a) }{1 - \cot(a) } = \sin(a) + \cos(a)$
​

Answer 1

Proof :

LHS

$\sf \dfrac{\cos(a)}{1 - \tan(a)} + \dfrac{\sin(a)}{1 - \cot(a)} \\\\ \sf = \dfrac{\cos(a)}{1 - \dfrac{\sin(a)}{\cos(a)}} + \dfrac{\sin(a)}{1 - \dfrac{\cos(a)}{\sin(a)}} \\\\ \sf = \dfrac{\cos(a)}{\dfrac{\cos(a) - \sin(a)}{\cos(a)}} + \dfrac{\sin(a)}{\dfrac{\sin(a) - cos(a)}{\sin(a)}} \\\\ \sf = \dfrac{\cos^{2}(a)}{\cos(a) - \sin(a)} + \dfrac{\sin^{2}(a)}{\sin(a) - \cos(a)} \\\\ \sf = \dfrac{\cos^{2}(a)}{\cos(a) - \sin(a)} - \dfrac{\sin^{2}}{\cos(a) - \sin(a)} \\\\ \sf = \dfrac{\cos^{2}(a) - \sin^{2}(a)}{\cos(a) - \sin(a)} \\\\ \sf = \dfrac{ \{\cos(a)- \sin(a)\}\{\cos(a) + \sin(a) \}}{\sin(a) - \cos(a)} \\\\ \sf = \cos(a) + \sin(a) \\\\ \sf = \sin(a) + \cos(a) \\\\ \sf = RHS$

Some Trigonometric Formulae :

$\sf\bullet \: \: \sin^{2}A + \cos^{2}A = 1 \\\\ \sf \bullet \: \: 1 + \tan^{2}A = \sec^{2}A \\\\ \sf \bullet \: \: \cot^{2} A + 1 = \csc^{2}A$

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

Answer 2

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TO PROVE :-

$\displaystyle{\sf{\frac{cos \theta}{1-tan\theta}+\frac{sin\theta}{1-cot\theta}=sin\theta+cos\theta}}$

[Consider angle (a) as θ]

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PROOF :-

LHS :-

$\displaystyle{\sf{\frac{cos\theta}{1-tan\theta}+\frac{sin\theta}{1-cot\theta}}}\\\\\\\displaystyle{\sf{=\frac{cos\theta}{1-\frac{sin\theta}{cos\theta}}+\frac{sin\theta}{1-\frac{cos\theta}{sin\theta}}}}\\\\\\\displaystyle{\sf{=\frac{cos\theta}{\frac{cos\theta-sin\theta}{cos\theta} } +\frac{sin\theta}{\frac{sin\theta-\cos\theta}{sin\theta} } }}$

$\displaystyle{\sf{=\frac{(cos\theta)^{2}}{cos\theta-sin\theta}+\frac{(sin\theta)^{2}}{sin\theta-cos\theta}}}\\\\\\\displaystyle{\sf{=\frac{(cos\theta)^{2}}{cos\theta-sin\theta}-\frac{(sin\theta)^{2}}{cos\theta-sin\theta}}}$

$\displaystyle{\sf{=\frac{cos^2\theta-sin^2\theta}{cos\theta-sin\theta} }}\\\\\\\displaystyle{\sf{=\frac{(cos\theta+sin\theta)(cos\theta-sin\theta)}{(cos\theta-sin\theta)} }}\\\\\\\displaystyle{\sf{[(cos\theta-sin\theta)\ gets\ cancelled\ in\ the\ numerator}}\\\displaystyle{\sf{and\ the\ denominator.]}}\\\\\displaystyle{\sf{=cos\theta+sin\theta} =\boxed{RHS}}}$

PROVED!

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Identities/formulae used :-

• a² - b² = (a + b)(a - b)
• tanθ = sinθ/cosθ
• cotθ = cosθ/sinθ