Question

Prove:

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

[Hint: Write the expression in terms of:
$\sf{sin \theta \:and \:cos \theta}$]

Chapter: Introduction to Trigonometry

Class: 10th

Concept: Trigonometric Identities.

• Quality answer needed!

• No spamming

• Thanks for helping me!​

Answer 1

$\large \pmb{\mathfrak{Question:-}}$

Prove that :

$\red{\tt\cfrac{tan\theta}{1-cot\theta}+\cfrac{cot\theta}{1-tan\theta}=1+sec\theta cosec\theta}$

⇒Hint : Write the expression in terms of $\tt sin\theta$ and $\tt cos\theta$.

$\large \pmb{\mathfrak{Solution:-}}$

$\red{\tt :\implies LHS= \cfrac{\cfrac{sin\theta}{cos\theta}}{1-\cfrac{cos\theta}{\sin\theta}}+\cfrac{\cfrac{cos\theta}{sin\theta}}{1-\cfrac{sin\theta}{cos\theta}} }$

$\green{\tt :\implies\cfrac{\cfrac{sin\theta}{cos\theta}}{\cfrac{sin\theta-cos\theta}{sin\theta}} +\cfrac{\cfrac{cos\theta}{sin\theta}}{\cfrac{cos\theta-sin\theta}{cos\theta}}}$

$\red{\tt:\implies \cfrac{sin\theta}{cos\theta}\times(\cfrac{sin\theta}{sin\theta-cos\theta})+\cfrac{cos\theta}{sin\theta}\times(\cfrac{cos\theta}{cos\theta-sin\theta})}$

$\green{\tt:\implies \cfrac{sin^2\theta}{cos\theta(sin\theta-cos\theta)}+\cfrac{cos^2\theta}{sin\theta(cos\theta-sin\theta)}}$

$\red{\tt:\implies Multiplying~\cfrac{cos^2\theta}{sin\theta(cos\theta-sin\theta)}~with~"-" ~:}$

$\green{\tt:\implies \cfrac{sin^2\theta}{cos\theta(sin\theta-cos\theta)}-\cfrac{cos^2\theta}{sin\theta(sin\theta-cos\theta)}}$

$\red{\tt :\implies Taking~LCM~and~solving:}$

$\green{\tt:\implies (\cfrac{sin^2\theta}{cos\theta(sin\theta-cos\theta)}\times\cfrac{sin\theta}{sin\theta}) -(\cfrac{cos^2\theta}{sin\theta(sin\theta-cos\theta)}\times \cfrac{cos\theta}{cos\theta})}$

$\red{\tt :\implies \cfrac{sin^3\theta-cos^3\theta}{sin\theta~cos\theta(sin\theta-cos\theta)}}$

$\green{\tt :\implies \cfrac{\cancel{(sin\theta-cos\theta)}(sin^2\theta+cos^2\theta+sin\theta~cos\theta)}{sin\theta~cos\theta\cancel{(sin\theta-cos\theta)}} }$

$\red{\tt:\implies \cfrac{sin^2\theta +cos^2\theta +sin\theta~cos\theta}{sin\theta~cos\theta}}$

$\green{\tt:\implies \cfrac{1+sin\theta~cos\theta}{sin\theta~cos\theta} }$

$\red{\tt:\implies \cfrac{1}{sin\theta~cos\theta}+\cancel{\cfrac{sin\theta~cos\theta}{sin\theta~cos\theta}}}$

$\green{\tt:\implies 1+\cfrac{1}{sin\theta}~\cfrac{1}{cos\theta}}$

$\red{\tt:\implies 1+sec\theta~cosec\theta=RHS}$

$\large \pmb{\mathfrak{Concepts~used:-}}$

$\red{\tt:\implies tan\theta=\cfrac{sin\theta}{cos\theta}}$

$\green{\tt:\implies cot\theta=\cfrac{cos\theta}{sin\theta}}$

$\red{\tt:\implies sin^2\theta+cos^2\theta=1 }$

$\green{\tt:\implies x^3-y^3=(x-y)(x^2+y^2+xy)}$

Answer 2

$\sf\underline\red{Question:-}$

Prove :

${\tt\cfrac{tan\theta}{1-cot\theta}+\cfrac{cot\theta}{1-tan\theta}=1+sec\theta cosec\theta}$

concept used :

${\tt:\implies tan\theta=\cfrac{sin\theta}{cos\theta}}$

${\tt:\implies cot\theta=\cfrac{cos\theta}{sin\theta}}$

${\tt:\implies sin^2\theta+cos^2\theta=1 }$

${\tt:\implies x^3-y^3=(x-y)(x^2+y^2+xy)}$

$\sf\underline\pink{Solution:-}$

${\tt :\implies LHS= \cfrac{\cfrac{sin\theta}{cos\theta}}{1-\cfrac{cos\theta}{\sin\theta}}+\cfrac{\cfrac{cos\theta}{sin\theta}}{1-\cfrac{sin\theta}{cos\theta}} }$

${\tt :\implies\cfrac{\cfrac{sin\theta}{cos\theta}}{\cfrac{sin\theta-cos\theta}{sin\theta}} +\cfrac{\cfrac{cos\theta}{sin\theta}}{\cfrac{cos\theta-sin\theta}{cos\theta}}}$

${\tt:\implies \cfrac{sin\theta}{cos\theta}\times(\cfrac{sin\theta}{sin\theta-cos\theta})+\cfrac{cos\theta}{sin\theta}\times(\cfrac{cos\theta}{cos\theta-sin\theta})}$

${\tt:\implies \cfrac{sin^2\theta}{cos\theta(sin\theta-cos\theta)}+\cfrac{cos^2\theta}{sin\theta(cos\theta-sin\theta)}}$

${\tt:\implies Multiplying~\cfrac{cos^2\theta}{sin\theta(cos\theta-sin\theta)}~with~"-" ~:}$

${\tt:\implies \cfrac{sin^2\theta}{cos\theta(sin\theta-cos\theta)}-\cfrac{cos^2\theta}{sin\theta(sin\theta-cos\theta)}}$

${\tt :\implies Taking~LCM~and~solving:}$

${\tt:\implies (\cfrac{sin^2\theta}{cos\theta(sin\theta-cos\theta)}\times\cfrac{sin\theta}{sin\theta}) -(\cfrac{cos^2\theta}{sin\theta(sin\theta-cos\theta)}\times \cfrac{cos\theta}{cos\theta})}$

${\tt :\implies \cfrac{sin^3\theta-cos^3\theta}{sin\theta~cos\theta(sin\theta-cos\theta)}}$

${\tt :\implies \cfrac{\cancel{(sin\theta-cos\theta)}(sin^2\theta+cos^2\theta+sin\theta~cos\theta)}{sin\theta~cos\theta\cancel{(sin\theta-cos\theta)}} }$

${\tt:\implies \cfrac{sin^2\theta +cos^2\theta +sin\theta~cos\theta}{sin\theta~cos\theta}}$

${\tt:\implies \cfrac{1+sin\theta~cos\theta}{sin\theta~cos\theta} }$

${\tt:\implies \cfrac{1}{sin\theta~cos\theta}+\cancel{\cfrac{sin\theta~cos\theta}{sin\theta~cos\theta}}}$

${\tt:\implies 1+\cfrac{1}{sin\theta}~\cfrac{1}{cos\theta}}$

${\tt:\implies 1+sec\theta~cosec\theta=RHS}$