Question

Class 10
Trigonometry


Please write the answers step by step
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Answer 1

To prove

$\star \tt \: \frac{tan \: \theta}{1 - cot \: \theta} + \frac{cot \: \theta }{1 - tan \: \theta } = 1 + cot \: \theta\: + tan \: \theta$

Solution:-

$\star \tt \: \frac{tan \: \theta}{1 - cot \: \theta } + \frac{cot \: \theta }{1 - tan \: \theta } = 1 + cot \: \theta \: + tan \: \theta$

Take L.H.S

$\implies \tt \: \frac{tan \: \theta}{1 - cot \: \theta } + \frac{cot \: \theta }{1 - tan\: \theta }$

$\implies \tt \: \frac{tan \: \theta}{1 - \frac{1}{tan \: \theta} } + \frac{cot \: \theta}{1 - tan \: \theta}$

$\implies \tt \: \frac{tan \: \theta }{(1 - tan \: \theta) } + \frac{cot \: \theta }{1 - tan\: \theta}$

$\implies \tt \: \frac{1}{1 - tan \: \theta} [ \frac{1}{tan \: \theta } - {tan}^{2} \theta )]$

$\implies \tt \: \frac{1}{1 - tan \: \theta} [ \frac{1 - {tan}^{3} \theta }{tan \: \theta } ]$

$\implies \tt \: \frac{1}{1 - tan \: \theta} [\frac{(1 - tan \: \theta )(1 + tan \: \theta+ {tan}^{2} \theta}{tan \: \theta} ] \: \:(Because\: {a}^{3} - {b}^{3} = (a - b)( {a}^{2} + ab + {b}^{2} )$

$\implies \tt \frac{1 + tan \: \theta+ {tan}^{2} \theta }{tan \: \theta}$

$\implies \tt \: cot \: \theta \: + 1 + tan \: \theta$

L.H.S = R.H.S

Hence, Proved