Question

$\frac{tanq}{1 - cotq} + \frac{cotq}{1 - tanq} = 1 + secqcosecq$
​

Answer 1

Solution:-

$\implies \rm \: \dfrac{ \tan(q) }{1 - \cot(q) } + \dfrac{ \cot(q) }{1 - \tan(q) }$

$\rm \implies \: \dfrac{ \tan(q) }{1 - \cot(q) } + \dfrac{ \cot(q ) }{1 - \dfrac{1}{ \cot(q) } }$

$\rm \implies \: \dfrac{ \dfrac{1}{ \cot(q) } }{1 - \cot(q) } - \dfrac{ \cot {}^{2} (q) }{1 - \cot(q) }$

$\rm \implies \: \dfrac{ \dfrac{1}{ \cot(q) } - \cot {}^{2} (q) }{1 - \cot(q) }$

$\rm \implies \: \dfrac{1 - \cot {}^{3} (q) }{ \cot(q) (1 - \cot(q) )}$

$\rm \implies \dfrac{(1 - \cot(q) )(1 + \cot(q) + \cot {}^{2} (q) )}{ \cot(q) (1 - \cot(q) )}$

$\rm \implies \dfrac{ \cancel{(1 - \cot(q) })(1 + \cot(q) + \cot {}^{2} (q) )}{ \cot(q) \cancel{ (1 - \cot(q)} )}$

$\rm \implies \dfrac{1 + \cot(q) + \cot {}^{2} (q) }{ \cot(q) }$

$\rm \: \implies \dfrac{ \cot(q) + \csc {}^{2} (q) }{ \cot(q) }$

$\rm \implies \: \dfrac{ \cot(q) }{ \cot(q) } + \dfrac{ \csc {}^{2} (q) }{ \cot(q) }$

$\rm \implies \: 1 + \dfrac{ \dfrac{1}{ \sin {}^{2} (q) } }{ \dfrac{ \cos(q) }{ \sin(q) } }$

$\rm \implies \: 1 + \dfrac{1}{ \sin {}^{2} (q) } \times \dfrac{ \sin(q) }{ \cos(q) }$

$\rm \implies \: 1 + \dfrac{1}{ \sin(q) \cos(q) }$

$\rm \implies \: 1 + \sec(q) \csc(q)$

hence proved

Answer 2

Solution :

$\bf{\dfrac{tanq}{1 - cotq} + \dfrac{cotq}{1 - tanq} = 1 + secqcosecq}$

From the above equation , we get :

• LHS = $\bf{\dfrac{tanq}{1 - cotq} + \dfrac{cotq}{1 - tanq}}$

• RHS = $\bf{1 + secqcosecq}$

By solving the LHS , we get :

$:\implies \bf{\dfrac{tanq}{1 - cotq} + \dfrac{cotq}{1 - tanq}}$

We know that,

• $\bf{cot\:\theta = \dfrac{cos\theta}{sin\theta}}$

• $\bf{cot\:\theta = \dfrac{cos\theta}{sin\theta}}$

By substituting them in the equation , we get :

$:\implies \bf{\dfrac{\dfrac{sinq}{cosq}}{1 - \dfrac{cosq}{sinq}} + \dfrac{\dfrac{\dfrac{cosq}{sinq}}{sinq}}{1 - \dfrac{sinq}{cosq}}}$

$:\implies \bf{\dfrac{\dfrac{sinq}{cosq}}{\dfrac{sinq - cosq}{sinq}} + \dfrac{\dfrac{cosq}{sinq}}{\dfrac{cosq - sinq}{cosq}}}$

$:\implies \bf{\dfrac{sinq}{cosq} \times \dfrac{sinq}{sinq - cosq} + \dfrac{cosq}{sinq} \times \dfrac{cosq}{cosq - sinq}}$

$:\implies \bf{\dfrac{sin^{2}q}{cosq(sinq - cosq)} + \dfrac{cos^{2}q}{sinq(cosq - sinq)}}$

$:\implies \bf{\dfrac{sin^{2}q}{cosq(sinq - cosq)} + \dfrac{cos^{2}q}{-sinq(sinq - cosq)}}$

$:\implies \bf{\dfrac{sin^{2}q}{cosq(sinq - cosq)} - \dfrac{cos^{2}q}{sinq(sinq - cosq)}}$

$:\implies \bf{\dfrac{sin^{2}q \times sinq - cos^{2}q \times cosq}{cosqsinq(sinq - cosq)}}$

$:\implies \bf{\dfrac{sin^{3}q - cos^{3}q}{cosqsinq(sinq - cosq)}}$

$:\implies \bf{\dfrac{sin^{3}q - cos^{3}q}{cosqsinq(sinq - cosq)}}$

By using the identity, we get :

$\bf{a^{3} - b^{3} = (a + b)(a^{2} + ab + b^{2}}$

$:\implies \bf{\dfrac{(sinq - cosq)(sin^{2}q + sinqcosq + cos^{q})}{cosqsinq(sinq - cosq)}}$

$:\implies \bf{\dfrac{(sinq - cosq)(sin^{2}q + sinqcosq + cos^{2}q)}{cosqsinq(sinq - cosq)}}$

We know that ,

$\bf{sin^{2}\theta + cos^{2}\theta = 1}$

By substituting it in the equation , we get :

$:\implies \bf{\dfrac{(sinq - cosq)(1 + sinqcosq)}{cosqsinq(sinq - cosq)}}$

$:\implies \bf{\dfrac{1 + sinqcosq}{cosqsinq}}$

$:\implies \bf{\dfrac{1 + sinqcosq}{cosqsinq}}$

We know that,

• $\bf{cos\:\theta = \dfrac{1}{sec\theta}}$

• $\bf{sin\:\theta = \dfrac{1}{cosec\theta}}$

By substituting them in the equation , we get :

$:\implies \bf{\dfrac{1 + \dfrac{1}{cosecq}\dfrac{1}{secq}}{\dfrac{1}{cosecq}\dfrac{1}{secq}}}$

$:\implies \bf{\dfrac{\dfrac{cosecqsecq + 1}{cosecqsecq}}{\dfrac{1}{cosecqsecq}}}$

$:\implies \bf{\dfrac{cosecqsecq + 1}{cosecqsecq} \times cosecqsecq}$

$:\implies \bf{cosecqsecq + 1}$

$\boxed{\therefore \bf{\dfrac{tanq}{1 - cotq} + \dfrac{cotq}{1 - tanq} = 1 + cosecqsecq}}$

Hence the LHS is 1 + cosecqsecq.

By putting the LHS and RHS together, we get :

$:\implies \bf{1 + cosecqsecq = 1 + secqcosecq}$

$\boxed{\therefore \bf{\dfrac{tanq}{1 - cotq} + \dfrac{cotq}{1 - tanq} = 1 + secqcosecq}}$

Hence proved !!