Question

Prove that:-

$\frac{tan \theta}{ 1-cot \theta} + \frac{ cot \theta}{ 1-tan \theta} =1 + sec \theta \: cosec \theta$

Class:- 10
Chapter:- Introduction to Trigonometry​

Answer 1

Solution :

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

$= \frac{{ \tan}^{2} \theta}{tan \theta \: - 1} + \: \frac{1}{ \tan \theta \: (1 \: - \: tan \theta)} \\ \\ \: = \frac{{ \tan}^{2} \theta}{tan \theta \: - 1} - \: \: \frac{1}{ \tan \theta \: ( tan \theta \: - \: 1)}$

$= \frac{ {tan}^{3} \theta \: - 1}{tan \theta(tan \theta - 1)} \\ \\ = \frac{(tan \theta \: - 1)( {tan}^{2} \theta \ + \: tan \theta) }{tan \theta \: (tan \theta - 1)}$

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

$= \frac{{tan}^{2} \theta \: + \: 1 \: + \: tan \theta}{tan \theta} \: \\ \\ = tan \theta \: + \: cot \theta \: + \: 1$

$= \frac{sin \theta}{cos \theta} + \frac{cos \theta}{sin \theta} + \: 1 \\ \\ = \frac{ {sin}^{2} \theta \: + \: {cos}^{2} \theta }{sin \theta \: cos \theta} + \: 1 \\ \\ = \frac{1}{sin \theta \: cos \theta} + \: 1 \\ \\ \: {sin}^{2} \theta \: + {cos}^{2} \theta \: = \: 1 \\ \\ = \: cosec \theta \: sec \theta \: + \: 1 \\ \\ = 1 \: + \: sec \theta \: cosec \theta \\ \\ = RHS \: \: \: Hence \: proved$

Answer 2

♣ Qᴜᴇꜱᴛɪᴏɴ :

$\bf{Prove\:\:\dfrac{tan\theta }{1-cot\theta }+\dfrac{\cot \theta }{1-tan\theta }=1+sec\theta \:cosec\theta}$

♣ ᴀɴꜱᴡᴇʀ :

$\sf{Manipulating\:left\:side}$

$\sf{\dfrac{\tan \left(\theta\right)}{1-\cot \left(\theta\right)}+\dfrac{\cot \left(\theta\right)}{1-\tan \left(\theta\right)}}$

Express with sin,cos

$\sf{\dfrac{\cot \left(\theta\right)}{1-\tan \left(\theta\right)}+\dfrac{\tan \left(\theta\right)}{1-\cot \left(\theta\right)}}$

Using the Basic Trigonometric identity: cot(x)= cos(x)/sin(x)

$\sf{\cot \left(\theta\right)=\dfrac{\cos \left(\theta\right)}{\sin \left(\theta\right)}}$

$\dfrac{\dfrac{\cos \left(\theta\right)}{\sin \left(\theta\right)}}{1-\tan \left(\theta\right)}+\dfrac{\tan \left(\theta\right)}{1-\dfrac{\cos \left(\theta\right)}{\sin \left(\theta\right)}}$

Using the Basic Trigonometric identity: tan(x)= sin(x)/cos(x)

$\tan \left(\theta\right)=\dfrac{\sin \left(\theta\right)}{\cos \left(\theta\right)}$

$=\dfrac{\dfrac{\cos \left(\theta\right)}{\sin \left(\theta\right)}}{1-\dfrac{\sin \left(\theta\right)}{\cos \left(\theta\right)}}+\dfrac{\dfrac{\sin \left(\theta\right)}{\cos \left(\theta\right)}}{1-\dfrac{\cos \left(\theta\right)}{\sin \left(\theta\right)}}$

$\bf{\bigstar \:\:Simplify\:\::\dfrac{\dfrac{\cos \left(\theta\right)}{\sin \left(\theta\right)}}{1-\dfrac{\sin \left(\theta\right)}{\cos \left(\theta\right)}}+\dfrac{\dfrac{\sin \left(\theta\right)}{\cos \left(\theta\right)}}{1-\dfrac{\cos \left(\theta\right)}{\sin \left(\theta\right)}}}$

$=\dfrac{\sin ^2\left(\theta\right)+\cos ^2\left(\theta\right)+\sin \left(\theta\right)\cos \left(\theta\right)}{\cos \left(\theta\right)\sin \left(\theta\right)}$

$=\dfrac{\cos ^2\left(\theta\right)+\sin ^2\left(\theta\right)+\cos \left(\theta\right)\sin \left(\theta\right)}{\cos \left(\theta\right)\sin \left(\theta\right)}$

$\mathrm{Use\:the\:following\:identity}:\quad \cos ^2\left(x\right)+\sin ^2\left(x\right)=1$

$=\dfrac{1+\cos \left(\theta\right)\sin \left(\theta\right)}{\cos \left(\theta\right)\sin \left(\theta\right)}$

$\mathrm{Manipulating\:right\:side}$

$1+\sec \left(\theta\right)\csc \left(\theta\right)$

Express with sin,cos

$=1+\dfrac{1}{\cos \left(\theta\right)}\cdot \dfrac{1}{\sin \left(\theta\right)}$

$\bf{\bigstar \:\:Simplify\:\::1+\dfrac{1}{\cos \left(\theta\right)}\cdot \dfrac{1}{\sin \left(\theta\right)}$

$=\dfrac{\cos \left(\theta\right)\sin \left(\theta\right)+1}{\cos \left(\theta\right)\sin \left(\theta\right)}$

$=\dfrac{1+\cos \left(\theta\right)\sin \left(\theta\right)}{\cos \left(\theta\right)\sin \left(\theta\right)}$

$\mathrm{We\:showed\:that\:the\:two\:sides\:could\:take\:the\:same\:form}$

$\Rightarrow \mathrm{True}$

$\boxed{\bf{\dfrac{tan\theta }{1-cot\theta }+\dfrac{\cot \theta }{1-tan\theta }=1+sec\theta \:cosec\theta}}$

Hence Proved !!!