Question

prove cos(90-theta)sec (90-theta)tan theta ÷cosec (90-theta)sin (90-theta)cot (90-theta) + tan (90-theta)÷cot theta = 2

Answer 1

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Answer 2

Answer and explanation:

To prove : $\frac{\cos(90-\theta)\sec (90-\theta)\tan \theta }{\csc (90-\theta)\sin (90-\theta)\cot (90-\theta)}+\frac{\tan (90-\theta)}{\cot \theta}=2$

Proof :

Taking LHS,

$=\frac{\cos(90-\theta)\sec (90-\theta)\tan \theta }{\csc (90-\theta)\sin (90-\theta)\cot (90-\theta)}+\frac{\tan (90-\theta)}{\cot \theta}$

Applying property of trigonometry,

$\sin (90-\theta)=\cos \theta\\\cos(90-\theta)=\sin\theta\\\sec (90-\theta)=\csc\theta\\\csc (90-\theta)=\sec\theta\\\cot (90-\theta)=\tan\theta\\\tan (90-\theta)=\cot\theta$

$=\frac{\sin\theta\csc\theta\tan \theta }{\sec\theta\cos\theta\tan\theta}+\frac{\cot\theta}{\cot \theta}$

We know, $\csc\tehta=\frac{1}{\sin\theta},\ \sec\tehta=\frac{1}{\cos\theta}$

$=\frac{\sin\theta\times \frac{1}{\sin\theta}\times \tan \theta }{\frac{1}{\cos\theta}\times \cos\theta\times \tan\theta}+\frac{\cot\theta}{\cot \theta}$

$=\frac{1}{1}+1$

$=1+1$

$=2$

LHS=RHS

Hence proved.