NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 8: Introduction to Trignometry and Its Applications - Exercise 8.3

Question 6

Step 1: Recall the trigonometric identities.

• \(\cot\alpha = \dfrac{\cos\alpha}{\sin\alpha} \implies \cot^2\alpha = \dfrac{\cos^2\alpha}{\sin^2\alpha}\)
• \(\csc\alpha = \dfrac{1}{\sin\alpha}\)

Step 2: Start with the left-hand side (LHS).

\[ LHS = 1 + \dfrac{\cot^2\alpha}{1 + \csc\alpha} \]

Step 3: Replace \(\cot^2\alpha\) and \(\csc\alpha\) with their \(\sin\alpha\) and \(\cos\alpha\) forms.

\[ LHS = 1 + \dfrac{\dfrac{\cos^2\alpha}{\sin^2\alpha}}{1 + \dfrac{1}{\sin\alpha}} \]

Step 4: Simplify the denominator \(1 + \dfrac{1}{\sin\alpha}\).

\[ 1 + \dfrac{1}{\sin\alpha} = \dfrac{\sin\alpha + 1}{\sin\alpha} \]

Step 5: Put this back into the fraction.

\[ LHS = 1 + \dfrac{\cos^2\alpha}{\sin^2\alpha} \times \dfrac{\sin\alpha}{\sin\alpha + 1} \]

Step 6: Simplify the multiplication.

\[ LHS = 1 + \dfrac{\cos^2\alpha}{\sin\alpha(\sin\alpha + 1)} \]

Step 7: Write both terms of LHS with a common denominator.

\[ LHS = \dfrac{\sin\alpha(\sin\alpha + 1)}{\sin\alpha(\sin\alpha + 1)} + \dfrac{\cos^2\alpha}{\sin\alpha(\sin\alpha + 1)} \]

\[ LHS = \dfrac{\sin^2\alpha + \sin\alpha + \cos^2\alpha}{\sin\alpha(\sin\alpha + 1)} \]

Step 8: Use the Pythagoras identity \(\sin^2\alpha + \cos^2\alpha = 1\).

\[ LHS = \dfrac{1 + \sin\alpha}{\sin\alpha(\sin\alpha + 1)} \]

Step 9: Cancel \((1+\sin\alpha)\) from numerator and denominator.

\[ LHS = \dfrac{1}{\sin\alpha} \]

Step 10: Recall that \(\dfrac{1}{\sin\alpha} = \csc\alpha\).

Therefore, LHS = RHS = \(\csc\alpha\). ✔