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

Question 4

Step 1: Write down what we need to prove.

We need to show that:

\[(\sin\alpha + \cos\alpha)(\tan\alpha + \cot\alpha) = \sec\alpha + \csc\alpha\]

Step 2: Expand the brackets on the left-hand side.

\((\sin\alpha+\cos\alpha)(\tan\alpha+\cot\alpha)\)

= \(\sin\alpha\tan\alpha + \sin\alpha\cot\alpha + \cos\alpha\tan\alpha + \cos\alpha\cot\alpha\)

Step 3: Write tan and cot in terms of sin and cos.

\(\tan\alpha = \dfrac{\sin\alpha}{\cos\alpha}, \; \cot\alpha = \dfrac{\cos\alpha}{\sin\alpha}\)

Step 4: Substitute these values.

= \(\sin\alpha \cdot \dfrac{\sin\alpha}{\cos\alpha} + \sin\alpha \cdot \dfrac{\cos\alpha}{\sin\alpha} + \cos\alpha \cdot \dfrac{\sin\alpha}{\cos\alpha} + \cos\alpha \cdot \dfrac{\cos\alpha}{\sin\alpha}\)

Step 5: Simplify each term carefully.

• \(\sin\alpha \cdot \dfrac{\sin\alpha}{\cos\alpha} = \dfrac{\sin^2\alpha}{\cos\alpha}\)
• \(\sin\alpha \cdot \dfrac{\cos\alpha}{\sin\alpha} = \cos\alpha\)
• \(\cos\alpha \cdot \dfrac{\sin\alpha}{\cos\alpha} = \sin\alpha\)
• \(\cos\alpha \cdot \dfrac{\cos\alpha}{\sin\alpha} = \dfrac{\cos^2\alpha}{\sin\alpha}\)

So the expression becomes:

\(= \dfrac{\sin^2\alpha}{\cos\alpha} + \cos\alpha + \sin\alpha + \dfrac{\cos^2\alpha}{\sin\alpha}\)

Step 6: Rearrange terms to group fractions together.

\(= \left(\dfrac{\sin^2\alpha}{\cos\alpha} + \cos\alpha\right) + \left(\dfrac{\cos^2\alpha}{\sin\alpha} + \sin\alpha\right)\)

Step 7: Simplify each group.

• \(\dfrac{\sin^2\alpha}{\cos\alpha} + \cos\alpha = \dfrac{\sin^2\alpha + \cos^2\alpha}{\cos\alpha} = \dfrac{1}{\cos\alpha} = \sec\alpha\)
• \(\dfrac{\cos^2\alpha}{\sin\alpha} + \sin\alpha = \dfrac{\cos^2\alpha + \sin^2\alpha}{\sin\alpha} = \dfrac{1}{\sin\alpha} = \csc\alpha\)

Step 8: Combine results.

= \(\sec\alpha + \csc\alpha\)

Therefore proved.