NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 8: Introduction to Trignometry and Its Applications - Exercise 8.1

Question 11

Step 1: Look at the first fraction: \[ \dfrac{\sin^2 22^\circ + \sin^2 68^\circ}{\cos^2 22^\circ + \cos^2 68^\circ} \]

Notice that \(68^\circ = 90^\circ - 22^\circ\). So, \(\sin 68^\circ = \cos 22^\circ\).

Therefore, numerator becomes: \(\sin^2 22^\circ + (\cos 22^\circ)^2 = \sin^2 22^\circ + \cos^2 22^\circ = 1\).

Similarly, denominator is: \(\cos^2 22^\circ + (\sin 22^\circ)^2 = 1\).

So the fraction = \(\dfrac{1}{1} = 1\).

Step 2: Now consider the term \(\cos 63^\circ \sin 27^\circ\).

Notice \(27^\circ = 90^\circ - 63^\circ\). So, \(\sin 27^\circ = \cos 63^\circ\).

Therefore, \(\cos 63^\circ \cdot \sin 27^\circ = \cos 63^\circ \cdot \cos 63^\circ = \cos^2 63^\circ\).

Step 3: Look at the other term: \(\sin^2 63^\circ\).

We know the identity: \(\sin^2 \theta + \cos^2 \theta = 1\).

So, \(\sin^2 63^\circ + \cos^2 63^\circ = 1\).

Step 4: Put everything together:

Whole expression = fraction (which is 1) + \(\sin^2 63^\circ + \cos^2 63^\circ\) (which is also 1).

So total = \(1 + 1 = 2\).

Final Answer: \(2\) (Option B)