Exercise 11.4 Solutions
From Class X Mathematics textbook by the State Council of Educational Research and Training, Telangana, Hyderabad
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:
\( = 2 + \cot \theta + \tan \theta – \sec \theta \csc \theta \)
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 \( \sin^2 \theta + \cos^2 \theta = 1 \):
\( = 2(1) = 2 \)
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 \( \cot \theta = \frac{1}{\tan \theta} \):
\( = \tan^2 \theta \cdot \frac{1}{\tan^2 \theta} = 1 \)
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.
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.
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.
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.
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 \( 1 – \sin^2 A = \cos^2 A \):
\( = \frac{\cos^2 A}{\cos^2 A} = 1 \)
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:
\( = 1 + 2 + 1 + \cot^2 A + 2 + 1 + \tan^2 A \)
\( = 7 + \tan^2 A + \cot^2 A \)
First two terms are difference of squares:
\( = (1 – \cos^2 \theta)(1 + \cot^2 \theta) \)
\( = \sin^2 \theta \cdot \csc^2 \theta \) (since \( 1 + \cot^2 \theta = \csc^2 \theta \))
\( = \sin^2 \theta \cdot \frac{1}{\sin^2 \theta} = 1 \)
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:
\( \sec \theta – \tan \theta = \frac{1}{p} \)
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, \( \cos \theta = \frac{\cot \theta}{\csc \theta} = \frac{\frac{k^2 – 1}{2k}}{\frac{k^2 + 1}{2k}} = \frac{k^2 – 1}{k^2 + 1} \)