Question

Sinø/1-cotø + cosø/1-tanø = sinø+cosø

Answer 1

ANSWER:

To Prove:

• sinø/1-cotø + cosø/1-tanø = sinø+cosø

Proof:

$:\longrightarrow\dfrac{\sin\theta}{1-\cot\theta}+\dfrac{\cos\theta}{1-\tan\theta}=\sin\theta+\cos\theta\\\\\text{Taking LHS,}\\\\:\implies\dfrac{\sin\theta}{1-\cot\theta}+\dfrac{\cos\theta}{1-\tan\theta}\\\\\text{We know that,}\\\\:\implies\cot\theta=\dfrac{\cos\theta}{\sin\theta}\:\:\&\:\:\tan\theta=\dfrac{\sin\theta}{\cos\theta}\\\\\text{So,}\\\\:\implies\dfrac{\sin\theta}{1-\dfrac{\cos\theta}{\sin\theta}}+\dfrac{\cos\theta}{1-\dfrac{\sin\theta}{\cos\theta}}\\\\:\implies\dfrac{\sin\theta}{\dfrac{\sin\theta-\cos\theta}{\sin\theta}}+\dfrac{\cos\theta}{\dfrac{\cos\theta-\sin\theta}{\cos\theta}}$

$:\implies\dfrac{\sin^2\theta}{\sin\theta-\cos\theta}+\dfrac{\cos^2\theta}{\cos\theta-\sin\theta}\\\\:\implies\dfrac{\sin^2\theta}{\sin\theta-\cos\theta}-\dfrac{\cos^2\theta}{\sin\theta-\cos\theta}\\\\:\implies\dfrac{\sin^2\theta-\cos^2\theta}{\sin\theta-\cos\theta}\\\\\text{We know that, a^2-b^2=(a+b)(a-b) . So,}\\\\:\implies\dfrac{(\sin\theta+\cos\theta)(\sin\theta-\cos\theta)}{\sin\theta-\cos\theta}\\\\\text{( \sin\theta-\cos\theta ) gets cut in numerator and denominator,}\\\\:\implies\sin\theta+\cos\theta \:\:=RHS\\\\\text{\bf{As, LHS = RHS}}\\\\\text{\bf{Hence Proved}}$

Formulae Used:

• tanø = sinø/cosø
• cotø = cosø/sinø
• a^2 - b^2 = (a + b)(a - b)