PYQ - Trigonometric identities — task. Maths CBSE Live product, Class 10 (2025-26).

Verify that

\frac{\tan θ}{1 - \cot θ} + \frac{\cot θ}{1 - \tan θ} = 1 + \sec θ cosec θ

Proof:

L.H.S \(=\)

\frac{\tan θ}{1 - \cot θ} + \frac{\cot θ}{1 - \tan θ}

\begin{array}{l} = \frac{\tan θ}{1 - \frac{1}{i θ}} + \frac{\frac{1}{i θ}}{1 - \tan θ} \\ = \frac{\tan θ}{\frac{\tan θ - 1}{i θ}} - \frac{\frac{1}{i θ}}{\tan θ - 1} \\ = \frac{i^{2} θ}{i θ - 1} - \frac{\frac{1}{i θ}}{i θ - 1} \\ = \frac{i^{2} θ - \frac{1}{i θ}}{\tan θ - 1} \\ = \frac{\tan^{3} θ - 1}{i θ (\tan θ - 1)} \\ = \frac{(i θ - 1) (i^{2} θ + \tan θ + i)}{i θ (\tan θ - 1)} \\ = \frac{i^{2} θ + \tan θ + 1}{i θ} \\ = \tan θ + 1 + \frac{1}{\tan θ} \\ = \tan θ + 1 + \cot θ \\ = \frac{i θ}{\cos θ} + 1 + \frac{i θ}{\sin θ} \\ = 1 + \frac{\sin^{2} θ + \cos^{2} θ}{i θ . \sin θ} \\ = 1 + \frac{i}{i θ . \sin θ} \\ = 1 + \frac{1}{i θ} \cdot \frac{1}{\sin θ} \\ = 1 + i θ . i θ \end{array}

\(=\) R.H.S

Hence the \(LHS=RHS\)

Did you find an error?

5. PYQ - Trigonometric identities