TBQ - Prove the given identity — task. Maths CBSE Live product, Class 10 (2026-27).

Prove that,

\frac{\sin θ - \cos θ + 1}{\sin θ + \cos θ - 1} = \frac{1}{\sec θ - \tan θ}

by using the trigonometric identity

\sec^{2} θ = 1 + \tan^{2} θ

Proof:

\(LHS=\)

\frac{\sin θ - \cos θ + 1}{\sin θ + \cos θ - 1}

Now, dividing both numerator and denominator by \(cos\ \theta\) we get,

\begin{array}{l} = \frac{(i θ + i θ) - 1}{(i θ - i θ) + 1} \\ Multiplying both numerator and denominator by (\tan θ - \sec θ) \\ = \frac{(i θ + i θ) - 1}{(i θ - i θ) + 1} \times \frac{(\tan θ - \sec θ)}{(\tan θ - \sec θ)} \\ = \frac{- i (1 + i θ - i θ)}{[(i θ - i θ) + 1] (\cot θ - cosec θ)} \\ = \frac{1}{i θ - i θ} \end{array}

\(=RHS\)

Hence proved.

Did you find an error?

7. TBQ - Prove the given identity