Question

Prove that:
cot theta+ cosec theta - 1/cot theta- cosec theta + 1= 1+cos theta/ sin theta​

Answer 1

Answer:

Step-by-step explanation:

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Cot theta + cosec theta-1/ cot theta-cosec theta+1= 1+cos theta/sin theta is proved

Solution:

We have to prove:-

Using trigonometric identity:

Applying this in L.H.S of equation we get,

Cancelling terms we get,

Hence proved

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Answer 2

Given:

$\frac{cotø + cosecø - 1}{cotø - cosecø + 1} = \frac{1 + cosø}{sinø}$

Solution:

$= > \frac{cotø + cosecø - 1}{cotø - cosecø + 1} = \frac{1 + cosø}{sinø} \\ = > \frac{ \frac{cosø}{sinø} + \frac{1}{sinø} - 1 }{ \frac{cosø}{sinø} - \frac{1}{sinø} + 1 } \\ = > \frac{ \frac{cosø + 1 - sinø}{sinø} }{ \frac{cosø - 1 + sinø}{sinø} } \\ = > \frac{cosø + (1 - sinø)}{cosø - (1 - sinø)} \\ = > \frac{cosø + (1 - sinø)}{cosø - (1 - sinø)} \times \frac{cosø + (1 - sinø)}{cosø + (1 - sinø)} \\ = > \frac{(cosø + (1 - sinø)) ^{2} }{cos ^{2} ø - (1 - sinø) ^{2} } \\ = > \frac{ {cos}^{2}ø + 2 \times cosø \times (1 - sinø) + (1 - sinø)^{2} }{ {cos}^{2} ø - (1 - 2sinø + {sin}^{2} ø)} \\ = > \frac{ {cos}^{2} ø + 2csø - 2sinøcosø + 1 - 2sinø + {sin}^{2} ø}{cos^{2}ø - 1 + 2sinø - {sin}^{2} ø }$

$= > \frac{ {sin}^{2} ø + {cos}^{2}ø - 2sinø + 2cos ø + 1 - 2sinøcosø}{1 - {sin}^{2}ø + 1 + 2sinø - {sin}^{2} ø } \\ = > \frac{1 + 1 - 2(sinø - cosø) - 2sinøcosø}{ - sin ^{2}ø + 2sinø } \\ = > \frac{2 - 2(sinø - cosø) - 2sinøcosø}{2(sinø - {sin}^{2}ø) } \\ = > \frac{2(1 - (sinø - cosø) - sinøcosø)}{2sinø(1 - sinø} \\ = > \frac{1 - sinø + cosø - sinøcosø}{sinø(1 - sinø)} \\ = > \frac{(1 - sinø) + cosø(1 - sinø)}{sinø(1 - sinø)} \\ = > \frac{(1 - sinø)(1 + cosø)}{sinø(1 - sinø)} \\ = > \: \frac{1 + cosø}{sinø} \\ = > L.H.S. = \: \frac{1 + cosø}{sinø} \\ = > R.H.S. = \: \frac{1 + cosø}{sinø} \\ L.H.S. = R.H.S \\Hence Proved$