Question

prove that (sinθ + cosθ)(cotθ + tanθ) = sec θ cosecθ​

Answer 1

ANSWER:

To Prove:

• (sinθ + cosθ)(cotθ + tanθ) = secθ+cosecθ

Proof:

$\text{We need to prove that,}\\\\:\longrightarrow(\sin\theta+\cos\theta)(\cot\theta+\tan\theta)=\sec\theta+\csc\theta\\\\\text{Solving LHS,}\\\\:\implies(\sin\theta+\cos\theta)(\cot\theta+\tan\theta)\\\\\text{We know that,}\\\\:\hookrightarrow\tan\theta=\dfrac{\sin\theta}{\cos\theta}\:\:\:and\:\:\:\cot\theta=\dfrac{\cos\theta}{\sin\theta}\\\\\text{So,}\\\\:\implies(\sin\theta+\cos\theta)\times(\cot\theta+\tan\theta)\\\\:\implies(\sin\theta+\cos\theta)\times\left(\dfrac{\cos\theta}{\sin\theta}+\dfrac{\sin\theta}{\cos\theta}\right)$

$:\implies\left(\sin\theta\times\dfrac{\cos\theta}{\sin\theta}\right)+\left(\sin\theta\times\dfrac{\sin\theta}{\cos\theta}\right)+\left(\cos\theta\times\dfrac{\cos\theta}{\sin\theta}\right)+\left(\cos\theta\times\dfrac{\sin\theta}{\cos\theta}\right)\\\\:\implies\cos\theta+\dfrac{\sin^2\theta}{\cos\theta}+\dfrac{\cos^2\theta}{\sin\theta}+\sin\theta\\\\\text{On taking LCM,}$

$:\implies\cos\theta+\dfrac{\sin^2\theta}{\cos\theta}+\dfrac{\cos^2\theta}{\sin\theta}+\sin\theta\\\\:\implies\dfrac{\sin\theta.\cos^2\theta+\sin^3\theta+\cos^3\theta+\sin^2\theta.\cos\theta}{\sin\theta.\cos\theta}\\\\:\implies\dfrac{\sin\theta(\cos^2\theta+\sin^2\theta)+\cos\theta(\cos^2\theta+\sin^2\theta)}{\sin\theta.\cos\theta}\\\\:\implies\dfrac{(\cos^2\theta+\sin^2\theta)(\sin\theta+\cos\theta)}{\sin\theta.\cos\theta}$

$\text{We know that,}\\\\:\hookrightarrow\sin^2\theta+\cos^2\theta=1\\\\\text{So,}\\\\:\implies\dfrac{(\cos^2\theta+\sin^2\theta)(\sin\theta+\cos\theta)}{\sin\theta.\cos\theta}\\\\:\implies\dfrac{(1)(\sin\theta+\cos\theta)}{\sin\theta.\cos\theta}\\\\:\implies\dfrac{\sin\theta+\cos\theta}{\sin\theta.\cos\theta}\\\\:\implies\dfrac{\sin\theta}{\sin\theta.\cos\theta}+\dfrac{\cos\theta}{\sin\theta.\cos\theta}\\\\:\implies\dfrac{1}{\cos\theta}+\dfrac{1}{\sin\theta}\\\\\text{We know that,}\\\\:\hookrightarrow\dfrac{1}{\sin\theta}=\csc\theta\:\:\:and\:\:\:\dfrac{1}{\cos\theta}=\sec\theta\\\\\text{So,}\\\\:\implies\dfrac{1}{\cos\theta}+\dfrac{1}{\sin\theta}\\\\\bf{:\implies\sec\theta+\csc\theta=RHS}\\\\\text{\bf{HENCE PROVED!!!}}$

Formulae Used:

$1.\tan\theta=\dfrac{\sin\theta}{\cos\theta}\\\\2.\cot\theta=\dfrac{\cos\theta}{\sin\theta}\\\\3.\sin^2\theta+\cos^2\theta=1\\\\4.\dfrac{1}{\sin\theta}=\csc\theta\\\\5.\dfrac{1}{\cos\theta}=\sec\theta$