Question

Prove using identities:-

$\dfrac{1 + {tan}^{2} \theta }{1 + {cot}^{2} \theta} ={ ( \dfrac{1 + tan \theta}{1 + cot \theta} ) }^{2} = {tan}^{2} \theta$
$\tt \red{!!!! \: \: No \: spam \: \: !!!! } \\ \sf{\underline{⋆ \: Proper \: explanation \: ⋆}}$

​

Answer 1

To Prove :

$\sf \dfrac{1 + {tan}^{2} \theta }{1 + {cot}^{2} \theta} ={ \Bigg( \dfrac{1 + tan \theta}{1 + cot \theta} \Bigg) }^{2} = {tan}^{2} \theta$

⠀ ⠀⠀ ⠀ ⠀⠀ ⠀

Proof :

⠀⠀ ⠀ ⠀⠀ ⠀

Type 1 :

$\sf \dfrac{1 + {tan}^{2} \theta }{1 + {cot}^{2} \theta}$

⠀⠀ ⠀ ⠀⠀ ⠀

We know that :

$\large \underline{\boxed{\bf{1 + tan^{2} = sec ^{2}}}}$

$\large \underline{\boxed{\bf{1 + cot^{2} = cosec ^{2}}}}$

$\sf : \implies \dfrac{sec^{2} \theta}{cosec^{2} \theta}$

⠀ ⠀⠀ ⠀

Now, we know that :

$\large \underline{\boxed{\bf{sec = \dfrac{1}{cos}}}}$

$\large \underline{\boxed{\bf{cosec = \dfrac{1}{sin}}}}$

$\sf : \implies \dfrac{\dfrac{1}{cos^{2} \theta}}{ \dfrac{1}{sin^{2} \theta}}$

$\sf : \implies \dfrac{1}{cos^{2} \theta}\times \dfrac{sin^{2} \theta}{1}$

$\sf : \implies \dfrac{sin^{2} \theta}{cos^{2} \theta}$

⠀ ⠀⠀ ⠀

Now, we know that :

$\large \underline{\boxed{\bf{\dfrac{sin}{cos} = tan}}}$

$\sf : \implies tan ^{2} \theta$

⠀⠀ ⠀ ⠀⠀ ⠀

Type 2 :

$\sf { \Bigg( \dfrac{1 + tan \theta}{1 + cot \theta} \Bigg) }^{2}$

⠀ ⠀⠀ ⠀

We know that :

$\large \underline{\boxed{\bf{tan = \dfrac{sin}{cos}}}}$

$\large \underline{\boxed{\bf{cot = \dfrac{cos}{sin}}}}$

$\sf : \implies \Bigg( \dfrac{1 + \dfrac{sin \theta}{cos\theta}}{1 + \dfrac{cos\theta}{sin \theta}}\Bigg)^{2}$

$\sf : \implies \Bigg( \dfrac{\dfrac{cos \theta + sin \theta}{cos\theta}}{ \dfrac{sin \theta + cos\theta}{sin \theta}}\Bigg)^{2}$

$\sf : \implies \Bigg( \dfrac{cos \theta + sin \theta}{cos\theta} \times \dfrac{sin \theta}{sin \theta + cos\theta}\Bigg)^{2}$

$\sf : \implies \Bigg( \dfrac{\cancel{cos \theta + sin \theta}}{cos\theta} \times \dfrac{sin \theta}{\cancel{cos \theta + sin\theta}}\Bigg)^{2}$

$\sf : \implies \Bigg( \dfrac{sin \theta}{cos\theta}\Bigg)^{2}$

⠀ ⠀⠀ ⠀

Now, we know that :

$\large \underline{\boxed{\bf{\dfrac{sin}{cos} = tan}}}$

$\sf : \implies (tan \theta)^{2}$

$\sf : \implies tan^{2} \theta$

⠀⠀ ⠀ ⠀⠀ ⠀

Type 3 :

$\sf {tan}^{2} \theta$

⠀⠀ ⠀ ⠀⠀ ⠀

As, Type 1 = Type 2 = Type 3 = tan²θ

Hence, proved.