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

Question 7

Step 1: Recall the co-function identities:

• \(\tan(90^\circ - \theta) = \cot\theta\)
• \(\sec(90^\circ - \theta) = \csc\theta\)

Step 2: Write the Left-Hand Side (LHS):

\(\tan\theta + \tan(90^\circ - \theta) = \tan\theta + \cot\theta\)

Step 3: Express \(\tan\theta\) and \(\cot\theta\) in terms of \(\sin\theta\) and \(\cos\theta\):

• \(\tan\theta = \dfrac{\sin\theta}{\cos\theta}\)
• \(\cot\theta = \dfrac{\cos\theta}{\sin\theta}\)

Step 4: Add them together:

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

Step 5: Use the Pythagorean identity:

\(\sin^2\theta + \cos^2\theta = 1\)

So, LHS = \(\dfrac{1}{\sin\theta \cos\theta}\)

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Step 6: Now take the Right-Hand Side (RHS):

\(\sec\theta \cdot \sec(90^\circ - \theta) = \sec\theta \cdot \csc\theta\)

Step 7: Write secant and cosecant in terms of sine and cosine:

• \(\sec\theta = \dfrac{1}{\cos\theta}\)
• \(\csc\theta = \dfrac{1}{\sin\theta}\)

Step 8: Multiply them:

\(\dfrac{1}{\cos\theta} \times \dfrac{1}{\sin\theta} = \dfrac{1}{\sin\theta \cos\theta}\)

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Step 9: Compare LHS and RHS:

LHS = RHS = \(\dfrac{1}{\sin\theta \cos\theta}\)

Final Result: The given identity is proved.