Question

Show that (1/cot theta)+ cot theta=sec theta×cosec theta

Answer 1

Question :–

▪︎ Show that : $\\ { \bold{ \dfrac{1}{ \cot( \theta) } + \cot( \theta) = \sec( \theta) . cosec( \theta)}}$

ANSWER :–

TO PROVE :–

$\\ { \bold{ \dfrac{1}{ \cot( \theta) } + \cot( \theta) = \sec( \theta) . cosec( \theta)}}$

SOLUTION :–

▪︎ Let's take L.H.S. –

$\\ { \bold{ = \dfrac{1}{ \cot( \theta) } + \cot( \theta) }}$

• We know that –

$\\ \implies{ \boxed{ \bold{ \cot( \theta) = { \frac{ \cos( \theta) }{ \sin( \theta) }} }}}$

• So that –

$\\ { \bold{ = \dfrac{1}{ \frac{ \cos( \theta) }{ \sin( \theta) } } + { \frac{ \cos( \theta) }{ \sin( \theta) }} }}$

$\\ { \bold{ = \frac{ \sin( \theta) }{ \cos( \theta) } + { \frac{ \cos( \theta) }{ \sin( \theta) }} }}$

$\\ { \bold{ = \frac{ \sin ^{2} ( \theta) + \cos^{2} ( \theta) }{ \sin( \theta) .\cos( \theta) } }}$

• We know that –

$\\ \to \: \: \: { \bold{ \sin ^{2} ( \theta) + \cos^{2} ( \theta) = 1 }}$

• So that –

$\\ { \bold{ = \frac{ 1 }{ \sin( \theta) .\cos( \theta) } }}$

• We know that –

$\\ \: \: \to{ \bold{ \frac{ 1 }{ \sin( \theta) } = cosec( \theta) \: \: and \: \: \frac{1}{ \cos( \theta) } = \sec( \theta) }}$

• So that –

$\\ { \bold{ = \sec( \theta) .cosec( \theta)}}$

$\\ { \bold{ = R.H.S. \: \: \: (Hence \: \: proved)}}$

Answer 2

$\bf To\ prove\ that:\ \sf \Big(\dfrac{1}{cot\ \theta}\Big)\ +\ cot\ \theta\ =\ sec\ \theta\ \times\ cosec\ \theta$

$\bf Proof:\\\\\\\tt Taking\ the\ Left\ Hand\ Side,\\\\\\\sf \Big(\dfrac{1}{cot\ \theta}\Big)\ +\ cot\ \theta\\\\\\\\\bf \Bigg[cot\ \theta\ =\ \dfrac{sin\ \theta}{cos\ \theta}\Bigg]\\\\\\\\\sf \Bigg(\dfrac{1}{\frac{cos\ \theta}{sin\ \theta}}\Bigg)\ +\ \dfrac{cos\ \theta}{sin\ \theta}\\\\\\\\\dfrac{sin\ \theta}{cos\ \theta}\ +\ \dfrac{cos\ \theta}{sin\ \theta}\\\\\\\\\tt Taking\ the\ LCM,\\\\\\\\\dfrac{sin^2\ \theta\ +\ cos^2\ \theta}{sin\theta\ cos\theta}$

$\bf \Bigg[sin^2\ \theta\ +\ cos^2\ \theta\ =\ 1\Bigg]\\\\\\\\\sf \dfrac{1}{sin\theta\ cos\theta}\\\\\\\\\bf \Bigg[\dfrac{1}{sin\ \theta}\ =\ cosec\ \theta;\ \dfrac{1}{cos\ \theta}\ =\ sec\ \theta\Bigg]\\\\\\\\sec\ \theta\ \times\ \cosec\ \theta\\\\\\\\=\ \tt Right\ Hand\ Side\\\\\\\\\bf Hence\ proved.$