Prove the following. [(tan^3(theta) – 1) / (tan(theta) – 1)] = sec^2(theta) + tan(theta)

Chapter 6 – Trigonometry – Text Book Solution

Problem set 6| Q 5.9| Page 138

Solution:-

[frac{tan^3 theta – 1}{tantheta – 1}]

[ = frac{left( tantheta – 1 right)left( tan^2 theta + tantheta times 1 + 1 right)}{tantheta – 1} left[ a^3 – b^3 = left( a – b right)left( a^2 + ab + b^2 right) right]]

[ = tan^2 theta + tantheta + 1]

[ = sec^2 theta + tantheta left( 1 + tan^2 theta = sec^2 theta right)]

Solution

To prove the given identity:

Let’s first simplify the left-hand side (LHS) of the equation:

LHS = (tan^3(theta) – 1) / (tan(theta) – 1) = [(tan(theta) – 1) (tan^2(theta) + tan(theta) + 1)] / (tan(theta) – 1) = tan^2(theta) + tan(theta) + 1

Now, let’s simplify the right-hand side (RHS) of the equation using the identity tan^2(theta) + 1 = sec^2(theta):

RHS = sec^2(theta) + tan(theta) = 1/cos^2(theta) + sin(theta)/cos(theta) = (1 + sin^2(theta)) / cos^2(theta) = (tan^2(theta) + 1) / cos^2(theta) = tan^2(theta) / cos^2(theta) + 1 / cos^2(theta) = sec^2(theta) + 1

We can see that LHS = RHS, and the identity is proven.

Chapter 6 – Trigonometry – Text Book Solution

Problem Set 6 |Q 5.9| P 138

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Q 5.9

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