NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 8: Introduction to Trignometry and Its Applications - Exercise 8.3
Question 11

Question. 11

Simplify \((1+\tan^2\theta)(1-\sin\theta)(1+\sin\theta)\).

Answer:

\(\displaystyle \sec^2\theta\,\cos^2\theta=1-\sin^2\theta=\cos^2\theta\) so the product equals \(1\).

Handwritten Notes

Video Explanation:

Detailed Answer with Explanation:

Step 1: Recall the trigonometric identity:

\[ 1 + \tan^2\theta = \sec^2\theta. \]

So we can replace \((1 + \tan^2\theta)\) with \(\sec^2\theta\).

Step 2: Look at the terms \((1 - \sin\theta)(1 + \sin\theta)\). This is in the form of \((a - b)(a + b) = a^2 - b^2.\)

Here, \(a = 1\) and \(b = \sin\theta\). So, \((1 - \sin\theta)(1 + \sin\theta) = 1^2 - (\sin\theta)^2 = 1 - \sin^2\theta.\)

Step 3: Another identity is:

\[ \sin^2\theta + \cos^2\theta = 1. \]

Rearranging gives: \(1 - \sin^2\theta = \cos^2\theta.\)

So, \((1 - \sin\theta)(1 + \sin\theta) = \cos^2\theta.\)

Step 4: Now substitute back into the expression:

\[ (1 + \tan^2\theta)(1 - \sin\theta)(1 + \sin\theta) = (\sec^2\theta)(\cos^2\theta). \]

Step 5: Recall the relation between secant and cosine:

\[ \sec\theta = \dfrac{1}{\cos\theta}. \]

So, \(\sec^2\theta = \dfrac{1}{\cos^2\theta}.\)

Step 6: Multiply:

\[ \sec^2\theta \cdot \cos^2\theta = \dfrac{1}{\cos^2\theta} \cdot \cos^2\theta. \]

Step 7: The \(\cos^2\theta\) in numerator and denominator cancel each other:

\[ = 1. \]

Final Answer: The value of the expression is \(1\).

NCERT Exemplar Solutions Class 10 – Mathematics – CHAPTER 8: Introduction to Trignometry and Its Applications – Exercise 8.3 | Detailed Answers