PYQ - Determine the following — task. Maths CBSE Live product, Class 10 (2025-26).

Check that

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

L.H.S \(=\)

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

Now, simplyfing this then we get,

\(LCM=\)

= \frac{\tan^{3} θ - 1}{\tan θ (\tan θ - 1)}

By applying the expansion of \(a^3-b^3\) then we get,

\(LHS=\)

By applying \(tan \theta= \frac{sin \theta}{cos \theta}\) , \(cot \theta = \frac{cos \theta}{sin \theta}\) and simplyfing then we get,

\(LHS=\)

Now, simplifying this then we get,

\(=\) \(1 + sec \theta \cdot cosec \theta\)

\(=\) R.H.S

Hence proved.

Answer variants:

= \frac{\tan^{2} θ}{\tan θ + 1} - \frac{\frac{1}{\tan θ}}{\tan θ + 1}

= \frac{(\tan θ - 1) (\tan^{2} θ + \tan θ + 1)}{\tan θ (\tan θ - 1)}

= \frac{(\tan θ + 1) (\tan^{2} θ + \tan θ + 1)}{\tan θ (\tan θ - 1)}

= 1 + \frac{\sin^{2} θ + \cos^{2} θ}{\cos θ \cdot \sin θ}

= \frac{\tan^{2} θ}{\tan θ - 1} - \frac{\frac{1}{\tan θ}}{\tan θ - 1}

Did you find an error?

6. PYQ - Determine the following