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

Question 14

Let us simplify step by step:

Step 1: Recall the trigonometric expansion formulas:

• \(\sin(A+B) = \sin A \cos B + \cos A \sin B\)
• \(\cos(A-B) = \cos A \cos B + \sin A \sin B\)

Step 2: Apply these formulas one by one.

For \(\sin(45^\circ+\theta)\):

\(\sin(45^\circ+\theta) = \sin 45^\circ \cos \theta + \cos 45^\circ \sin \theta\)

Step 3: We know that \(\sin 45^\circ = \cos 45^\circ = \dfrac{1}{\sqrt{2}}\).

So, \(\sin(45^\circ+\theta) = \dfrac{1}{\sqrt{2}}(\cos \theta + \sin \theta)\).

Step 4: Now expand \(\cos(45^\circ-\theta)\):

\(\cos(45^\circ-\theta) = \cos 45^\circ \cos \theta + \sin 45^\circ \sin \theta\)

Again, \(\cos 45^\circ = \sin 45^\circ = \dfrac{1}{\sqrt{2}}\).

So, \(\cos(45^\circ-\theta) = \dfrac{1}{\sqrt{2}}(\cos \theta + \sin \theta)\).

Step 5: Now subtract the two expressions:

\(\sin(45^\circ+\theta) - \cos(45^\circ-\theta) = \dfrac{1}{\sqrt{2}}(\cos \theta + \sin \theta) - \dfrac{1}{\sqrt{2}}(\cos \theta + \sin \theta)\)

Step 6: Since both terms are exactly the same, their difference is:

\(0\).

Final Answer: Option B (0).