CBSE Class 10 Mathematics Trigonometry Assignment Set 25

Very Short Answer Type Questions

Question 1. If \( \cos A = \frac{4}{5} \), then find the value of \( \tan A \).
Answer:
\( \cos A = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{4}{5} \). Let Base \( = 4k \) and Hypotenuse \( = 5k \).
Perpendicular \( = \sqrt{(5k)^2 - (4k)^2} = \sqrt{9k^2} = 3k \).
\( \tan A = \frac{\text{Perpendicular}}{\text{Base}} = \frac{3k}{4k} = \frac{3}{4} \).
In simple words: The sides are in the ratio 3 : 4 : 5, so the tangent is 3 over 4.

Teacher's Note:
1. Draw the right triangle first and mark the given sides.
2. Write the final answer in its simplest form.

Question 2. If \( \sqrt{3}\sin \theta - \cos \theta = 0 \) and \( 0^\circ < \theta < 90^\circ \), find the value of \( \theta \).
Answer:
\( \sqrt{3}\sin \theta = \cos \theta \Rightarrow \frac{\sin \theta}{\cos \theta} = \frac{1}{\sqrt{3}} \Rightarrow \tan \theta = \frac{1}{\sqrt{3}} = \tan 30^\circ \).
Therefore \( \theta = 30^\circ \).
In simple words: Divide both sides by cos θ to get tan θ, then read the angle from the table.

Teacher's Note:
1. Divide by cos θ (it is not zero for an acute angle).
2. Compare the value of tan θ with the standard table.

Question 3. In a triangle ABC, \( \angle B = 90^\circ \). If \( AB = 12 \) cm and \( AC = 13 \) cm, then find \( \sin A \) and \( \cos C \).
Answer:
\( BC = \sqrt{AC^2 - AB^2} = \sqrt{169 - 144} = 5 \) cm.
\( \sin A = \frac{BC}{AC} = \frac{5}{13} \) and \( \cos C = \frac{BC}{AC} = \frac{5}{13} \).
In simple words: Both ratios use the side BC, which is opposite to A and next to C, so they are equal.

Teacher's Note:
1. Find the third side with the Pythagoras theorem.
2. Note that sin A = cos C because A and C are complementary angles.

Question 4. Prove that: \( \frac{\tan \theta + \sin \theta}{\tan \theta - \sin \theta} = \frac{\sec \theta + 1}{\sec \theta - 1} \).
Answer:
LHS \( = \frac{\frac{\sin \theta}{\cos \theta} + \sin \theta}{\frac{\sin \theta}{\cos \theta} - \sin \theta} = \frac{\sin \theta\left( \frac{1}{\cos \theta} + 1 \right)}{\sin \theta\left( \frac{1}{\cos \theta} - 1 \right)} = \frac{\sec \theta + 1}{\sec \theta - 1} = \) RHS. Hence proved.
In simple words: Write tan θ as sin θ over cos θ and take sin θ out as a common factor.

Teacher's Note:
1. Take sin θ common from the numerator and the denominator.
2. Replace 1/cos θ by sec θ to reach the right side.

Question 5. Find the value of \( 2(\cos^2 45^\circ + \tan^2 60^\circ) - 6(\sin^2 45^\circ - \tan^2 30^\circ) \).
Answer:
\( = 2\left[ \left( \frac{1}{\sqrt{2}} \right)^2 + (\sqrt{3})^2 \right] - 6\left[ \left( \frac{1}{\sqrt{2}} \right)^2 - \left( \frac{1}{\sqrt{3}} \right)^2 \right] \)
\( = 2\left( \frac{1}{2} + 3 \right) - 6\left( \frac{1}{2} - \frac{1}{3} \right) = 2 \times \frac{7}{2} - 6 \times \frac{1}{6} = 7 - 1 = 6 \).
In simple words: Put in the table values of the squares and simplify the two brackets.

Teacher's Note:
1. Write each standard value inside the brackets.
2. Add and subtract fractions with a common denominator.

Question 6. If \( \sin A + \cos B = 1 \), \( A = 30^\circ \) and B is an acute angle, then find the value of B.
Answer:
\( \sin 30^\circ + \cos B = 1 \Rightarrow \frac{1}{2} + \cos B = 1 \Rightarrow \cos B = \frac{1}{2} = \cos 60^\circ \).
Therefore \( B = 60^\circ \).
In simple words: Substitute A = 30°, find cos B and read the acute angle from the table.

Teacher's Note:
1. Use sin 30° = 1/2.
2. Since B is acute, only one angle has cos B = 1/2.

Question 7. If \( \sin \alpha = \frac{\sqrt{3}}{2} \) and \( \cos \beta = 0 \), then find the value of \( \beta - \alpha \).
Answer:
\( \sin \alpha = \frac{\sqrt{3}}{2} = \sin 60^\circ \Rightarrow \alpha = 60^\circ \).
\( \cos \beta = 0 = \cos 90^\circ \Rightarrow \beta = 90^\circ \).
\( \beta - \alpha = 90^\circ - 60^\circ = 30^\circ \).
In simple words: Identify each angle from the table of standard values and subtract.

Teacher's Note:
1. Remember cos 90° = 0.
2. Subtract in the order asked: β − α.

Question 8. Evaluate: \( \sin^2 60^\circ + 2\tan 45^\circ - \cos^2 30^\circ \).
Answer:
\( \sin^2 60^\circ + 2\tan 45^\circ - \cos^2 30^\circ = \left( \frac{\sqrt{3}}{2} \right)^2 + 2(1) - \left( \frac{\sqrt{3}}{2} \right)^2 = 2 \).
In simple words: The first and the last term are equal, so they cancel and only 2 tan 45° = 2 remains.

Teacher's Note:
1. Note that sin 60° = cos 30°.
2. Cancel equal terms before calculating.

Question 9. Without using the trigonometric table, find the value of \( \frac{\cot 54^\circ}{\tan 36^\circ} + \frac{\tan 20^\circ}{\cot 70^\circ} - 2 \).
Answer:
\( \cot 54^\circ = \cot(90^\circ - 36^\circ) = \tan 36^\circ \) and \( \tan 20^\circ = \tan(90^\circ - 70^\circ) = \cot 70^\circ \).
\( \frac{\tan 36^\circ}{\tan 36^\circ} + \frac{\cot 70^\circ}{\cot 70^\circ} - 2 = 1 + 1 - 2 = 0 \).
In simple words: Change each ratio to its complementary angle; every fraction then becomes 1.

Teacher's Note:
1. Use cot (90° − θ) = tan θ and tan (90° − θ) = cot θ.
2. Look for pairs of angles that add up to 90°.

Question 10. Find the value of \( \left( \frac{\sin 35^\circ}{\cos 55^\circ} \right)^2 + \left( \frac{\cos 43^\circ}{\sin 47^\circ} \right)^2 - 2\cos 60^\circ \).
Answer:
\( \sin 35^\circ = \sin(90^\circ - 55^\circ) = \cos 55^\circ \) and \( \cos 43^\circ = \cos(90^\circ - 47^\circ) = \sin 47^\circ \).
\( \left( \frac{\cos 55^\circ}{\cos 55^\circ} \right)^2 + \left( \frac{\sin 47^\circ}{\sin 47^\circ} \right)^2 - 2 \times \frac{1}{2} = 1 + 1 - 1 = 1 \).
In simple words: Both fractions equal 1 because the angles are complementary, and 2 cos 60° = 1.

Teacher's Note:
1. Check that 35° + 55° = 90° and 43° + 47° = 90°.
2. Use cos 60° = 1/2.

Question 11. If \( \tan A = \cot B \), then find the value of \( (A + B) \).
Answer:
\( \cot B = \tan(90^\circ - B) \), so \( \tan A = \tan(90^\circ - B) \Rightarrow A = 90^\circ - B \).
Hence \( A + B = 90^\circ \).
In simple words: Equal tangents of acute angles mean the angles are equal, so A and B are complementary.

Teacher's Note:
1. Convert cot B to tan (90° − B).
2. This result is used often: tan A = cot B means A + B = 90°.

Question 12. Show that \( \sec^4 \theta - \sec^2 \theta = \tan^4 \theta + \tan^2 \theta \).
Answer:
LHS \( = \sec^2 \theta(\sec^2 \theta - 1) = (1 + \tan^2 \theta)(1 + \tan^2 \theta - 1) = (1 + \tan^2 \theta)\tan^2 \theta = \tan^2 \theta + \tan^4 \theta = \) RHS. Hence shown.
In simple words: Take sec² θ common and use sec² θ = 1 + tan² θ.

Teacher's Note:
1. Factor the left side as sec² θ (sec² θ − 1).
2. Replace sec² θ by 1 + tan² θ everywhere.

Question 13. If \( \tan A + \frac{1}{\tan A} = 2 \), find the value of \( \tan^{100} A + \frac{1}{\tan^{100} A} \).
Answer:
Let \( \tan A = x \). Then \( x + \frac{1}{x} = 2 \Rightarrow x^2 - 2x + 1 = 0 \Rightarrow (x - 1)^2 = 0 \Rightarrow x = 1 \).
So \( \tan A = 1 \) and \( \tan^{100} A + \frac{1}{\tan^{100} A} = 1 + 1 = 2 \).
In simple words: The only number that adds to its reciprocal to make 2 is 1, so every power of it is 1.

Teacher's Note:
1. Form a quadratic in tan A and solve it.
2. Any power of 1 is 1.

Question 14. If \( \sin \theta = 0 \), then find the value of \( (\sin^4 \theta + \cos^4 \theta) \).
Answer:
\( \sin \theta = 0 \Rightarrow \theta = 0^\circ \).
\( \sin^4 0^\circ + \cos^4 0^\circ = 0 + 1 = 1 \).
In simple words: At 0° the sine is 0 and the cosine is 1, so the sum of fourth powers is 1.

Teacher's Note:
1. Identify the angle first.
2. Remember cos 0° = 1.