10th Maths Trigonometry Exercise 11.4 Solutions

Exercise 11.4 Solutions

Exercise 11.4 Solutions

From Class X Mathematics textbook by the State Council of Educational Research and Training, Telangana, Hyderabad

1. Evaluate the following:
(i) \( (1 + \tan \theta + \sec \theta) (1 + \cot \theta – \csc \theta) \)

Let’s expand the expression:

\( = 1(1 + \cot \theta – \csc \theta) + \tan \theta(1 + \cot \theta – \csc \theta) + \sec \theta(1 + \cot \theta – \csc \theta) \)

\( = 1 + \cot \theta – \csc \theta + \tan \theta + \tan \theta \cot \theta – \tan \theta \csc \theta + \sec \theta + \sec \theta \cot \theta – \sec \theta \csc \theta \)

Simplify using trigonometric identities:

\( \tan \theta \cot \theta = 1 \)

\( \tan \theta \csc \theta = \frac{\sin \theta}{\cos \theta} \cdot \frac{1}{\sin \theta} = \frac{1}{\cos \theta} = \sec \theta \)

\( \sec \theta \cot \theta = \frac{1}{\cos \theta} \cdot \frac{\cos \theta}{\sin \theta} = \frac{1}{\sin \theta} = \csc \theta \)

Substituting back:

\( = 1 + \cot \theta – \csc \theta + \tan \theta + 1 – \sec \theta + \sec \theta + \csc \theta – \sec \theta \csc \theta \)

Many terms cancel out:

\( = 2 + \cot \theta + \tan \theta – \sec \theta \csc \theta \)

(ii) \( (\sin \theta + \cos \theta)^2 + (\sin \theta – \cos \theta)^2 \)

Expand both squares:

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

Combine like terms:

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

Factor out 2:

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

Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):

\( = 2(1) = 2 \)

(iii) \( (\sec^2 \theta – 1) (\csc^2 \theta – 1) \)

We know that:

\( \sec^2 \theta – 1 = \tan^2 \theta \)

\( \csc^2 \theta – 1 = \cot^2 \theta \)

So the expression becomes:

\( \tan^2 \theta \cdot \cot^2 \theta \)

Since \( \cot \theta = \frac{1}{\tan \theta} \):

\( = \tan^2 \theta \cdot \frac{1}{\tan^2 \theta} = 1 \)

2. Show that \( (\csc \theta – \cot \theta)^2 = \frac{1 – \cos \theta}{1 + \cos \theta} \)

Start with the left side:

\( (\csc \theta – \cot \theta)^2 \)

Express in terms of sine and cosine:

\( = \left( \frac{1}{\sin \theta} – \frac{\cos \theta}{\sin \theta} \right)^2 = \left( \frac{1 – \cos \theta}{\sin \theta} \right)^2 = \frac{(1 – \cos \theta)^2}{\sin^2 \theta} \)

Using \( \sin^2 \theta = 1 – \cos^2 \theta \):

\( = \frac{(1 – \cos \theta)^2}{(1 – \cos \theta)(1 + \cos \theta)} \)

Cancel \( (1 – \cos \theta) \):

\( = \frac{1 – \cos \theta}{1 + \cos \theta} \)

Which matches the right side.

3. Show that \( \sqrt{\frac{1 + \sin A}{1 – \sin A}} = \sec A + \tan A \)

Start with the left side:

Rationalize the numerator by multiplying numerator and denominator by \( 1 + \sin A \):

\( \sqrt{\frac{(1 + \sin A)^2}{1 – \sin^2 A}} = \sqrt{\frac{(1 + \sin A)^2}{\cos^2 A}} = \frac{1 + \sin A}{\cos A} \)

Split the fraction:

\( = \frac{1}{\cos A} + \frac{\sin A}{\cos A} = \sec A + \tan A \)

Which matches the right side.

4. Show that \( \frac{1 – \tan^2 A}{\cot^2 A – 1} = \tan^2 A \)

Start with the left side:

Express everything in terms of tan A:

\( \cot A = \frac{1}{\tan A} \), so \( \cot^2 A = \frac{1}{\tan^2 A} \)

Substitute:

\( \frac{1 – \tan^2 A}{\frac{1}{\tan^2 A} – 1} = \frac{1 – \tan^2 A}{\frac{1 – \tan^2 A}{\tan^2 A}} \)

Divide by a fraction is same as multiplying by its reciprocal:

\( = (1 – \tan^2 A) \cdot \frac{\tan^2 A}{1 – \tan^2 A} = \tan^2 A \)

Which matches the right side.

5. Show that \( \frac{1}{\cos \theta} – \cos \theta = \tan \theta \cdot \sin \theta \)

Start with the left side:

\( \frac{1}{\cos \theta} – \cos \theta = \frac{1 – \cos^2 \theta}{\cos \theta} = \frac{\sin^2 \theta}{\cos \theta} \)

Now, the right side:

\( \tan \theta \cdot \sin \theta = \frac{\sin \theta}{\cos \theta} \cdot \sin \theta = \frac{\sin^2 \theta}{\cos \theta} \)

Both sides are equal.

6. Simplify \( \sec A (1 – \sin A) (\sec A + \tan A) \)

First, expand the expression:

\( = \sec A (1 – \sin A) \sec A + \sec A (1 – \sin A) \tan A \)

\( = \sec^2 A (1 – \sin A) + \sec A \tan A (1 – \sin A) \)

Factor out \( (1 – \sin A) \):

\( = (1 – \sin A)(\sec^2 A + \sec A \tan A) \)

Express in terms of cosine and sine:

\( = (1 – \sin A)\left( \frac{1}{\cos^2 A} + \frac{\sin A}{\cos^2 A} \right) = (1 – \sin A)\left( \frac{1 + \sin A}{\cos^2 A} \right) \)

Numerator becomes \( 1 – \sin^2 A = \cos^2 A \):

\( = \frac{\cos^2 A}{\cos^2 A} = 1 \)

7. Prove that \( (\sin A + \csc A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A \)

Expand both squares:

\( = \sin^2 A + 2\sin A \csc A + \csc^2 A + \cos^2 A + 2\cos A \sec A + \sec^2 A \)

Simplify using identities:

\( \sin A \csc A = 1 \), \( \cos A \sec A = 1 \)

\( \csc^2 A = 1 + \cot^2 A \), \( \sec^2 A = 1 + \tan^2 A \)

Also, \( \sin^2 A + \cos^2 A = 1 \)

Substitute:

\( = 1 + 2(1) + (1 + \cot^2 A) + 2(1) + (1 + \tan^2 A) \)

Combine like terms:

\( = 1 + 2 + 1 + \cot^2 A + 2 + 1 + \tan^2 A \)

\( = 7 + \tan^2 A + \cot^2 A \)

8. Simplify \( (1 – \cos \theta) (1 + \cos \theta) (1 + \cot^2 \theta) \)

First two terms are difference of squares:

\( = (1 – \cos^2 \theta)(1 + \cot^2 \theta) \)

\( = \sin^2 \theta \cdot \csc^2 \theta \) (since \( 1 + \cot^2 \theta = \csc^2 \theta \))

\( = \sin^2 \theta \cdot \frac{1}{\sin^2 \theta} = 1 \)

9. If \( \sec \theta + \tan \theta = p \), then what is the value of \( \sec \theta – \tan \theta \)?

We know the identity:

\( \sec^2 \theta – \tan^2 \theta = 1 \)

This can be written as:

\( (\sec \theta + \tan \theta)(\sec \theta – \tan \theta) = 1 \)

Given \( \sec \theta + \tan \theta = p \), then:

\( p (\sec \theta – \tan \theta) = 1 \)

Therefore:

\( \sec \theta – \tan \theta = \frac{1}{p} \)

10. If \( \csc \theta + \cot \theta = k \), then prove that \( \cos \theta = \frac{k^2 – 1}{k^2 + 1} \)

We know the identity:

\( \csc^2 \theta – \cot^2 \theta = 1 \)

This can be written as:

\( (\csc \theta + \cot \theta)(\csc \theta – \cot \theta) = 1 \)

Given \( \csc \theta + \cot \theta = k \), then \( \csc \theta – \cot \theta = \frac{1}{k} \)

Add the two equations:

\( 2\csc \theta = k + \frac{1}{k} = \frac{k^2 + 1}{k} \)

Subtract the two equations:

\( 2\cot \theta = k – \frac{1}{k} = \frac{k^2 – 1}{k} \)

Now, \( \cos \theta = \frac{\cot \theta}{\csc \theta} = \frac{\frac{k^2 – 1}{2k}}{\frac{k^2 + 1}{2k}} = \frac{k^2 – 1}{k^2 + 1} \)

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