Question

show that or prove that question attached ​

Answer 1

Step-by-step explanation:

To Prove : ${\sf{\ \ {\dfrac{tan^3 \theta - 1}{tan \theta - 1}} = sec^2 \theta + tan \theta}}$

L.H.S. = ${\sf{\ \ {\dfrac{tan^3 \theta - 1}{tan \theta - 1}}}}$

→ ${\sf{ {\dfrac{ (tan \theta)^3 - (1)^3}{tan \theta - 1}}}}$

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

${\sf{\red{Here, \ a = tan \theta , \ b = 1}}}$

→ ${\sf{ {\dfrac{(tan \theta - 1)[ (tan \theta)^2 + (1)^2 + (tan \theta)(1)]}{tan \theta - 1}}}}$

→ ${\sf{ {\dfrac{{\cancel{(tan \theta - 1)}}[ (tan \theta)^2 + (1)^2 + (tan \theta)(1)]}{{\cancel{tan \theta - 1}}}}}}$

→ ${\sf{tan^2 \theta + 1 + tan \theta}}$

${\boxed{\sf{\red{Identity \ : \ tan^2 \theta + 1 = sec^2 \theta}}}}$

→ ${\sf{sec^2 \theta + tan \theta}}$

= R.H.S.

Hence, proved !!

${\rule{200}{2}}$

Similar Identities which are frequently used :

• sin²θ + cos²θ = 1

• cot²θ + 1 = cosec²θ