RS Aggarwal Class 10 Mathematics Solutions Chapter 7 Trigonometric Ratios of Complementary Angles

Question 1. Simplify the following.
(i) \( \frac{\sin 16°}{\cos 74°} \)
(ii) \( \frac{\sec 11°}{\cosec 79°} \)
(iii) \( \frac{\tan 27°}{\cot 63°} \)
(iv) \( \frac{\cos 35°}{\sin 55°} \)
(v) \( \frac{\cosec 42°}{\sec 48°} \)
(vi) \( \frac{\cot 38°}{\tan 52°} \)
Answer: Each expression equals 1.
(i) \( \frac{\sin 16°}{\cos 74°} = \frac{\sin (90° - 74°)}{\cos 74°} = \frac{\cos 74°}{\cos 74°} = 1 \)
(ii) \( \frac{\sec 11°}{\cosec 79°} = \frac{\sec (90° - 79°)}{\cosec 79°} = \frac{\cosec 79°}{\cosec 79°} = 1 \)
(iii) \( \frac{\tan 27°}{\cot 63°} = \frac{\tan (90° - 63°)}{\cot 63°} = \frac{\cot 63°}{\cot 63°} = 1 \)
(iv) \( \frac{\cos 35°}{\sin 55°} = \frac{\cos (90° - 55°)}{\sin 55°} = \frac{\sin 55°}{\sin 55°} = 1 \)
(v) \( \frac{\cosec 42°}{\sec 48°} = \frac{\cosec (90° - 48°)}{\sec 48°} = \frac{\sec 48°}{\sec 48°} = 1 \)
(vi) \( \frac{\cot 38°}{\tan 52°} = \frac{\cot (90° - 52°)}{\tan 52°} = \frac{\tan 52°}{\tan 52°} = 1 \)
In simple words: Complementary angle relationships help reduce each fraction to a single ratio divided by itself, which always gives 1.

Exam Tip: Remember that \( \sin(90° - \theta) = \cos \theta \), \( \sec(90° - \theta) = \cosec \theta \), and \( \tan(90° - \theta) = \cot \theta \). Use these identities to convert angles so numerator and denominator match.

Question 2. Prove the following.
(i) \( \cos 81° - \sin 9° = 0 \)
(ii) \( \tan 71° - \cot 19° = 0 \)
(iii) \( \cosec 80° - \sec 10° = 0 \)
(iv) \( \cosec^2 72° - \tan^2 18° = 1 \)
(v) \( \cos^2 75° + \cos^2 15° = 1 \)
(vi) \( \tan^2 66° - \cot^2 24° = 0 \)
Answer:
(i) LHS \( = \cos 81° - \sin 9° = \cos(90° - 9°) - \sin 9° = \sin 9° - \sin 9° = 0 = \) RHS
(ii) LHS \( = \tan 71° - \cot 19° = \tan(90° - 19°) - \cot 19° = \cot 19° - \cot 19° = 0 = \) RHS
(iii) LHS \( = \cosec 80° - \sec 10° = \cosec(90° - 10°) - \sec 10° = \sec 10° - \sec 10° = 0 = \) RHS
(iv) LHS \( = \cosec^2 72° - \tan^2 18° = \cosec^2(90° - 18°) - \tan^2 18° = \sec^2 18° - \tan^2 18° = 1 = \) RHS
(v) LHS \( = \cos^2 75° + \cos^2 15° = \cos^2(90° - 15°) + \cos^2 15° = \sin^2 15° + \cos^2 15° = 1 = \) RHS
(vi) LHS \( = \tan^2 66° - \cot^2 24° = \tan^2(90° - 24°) - \cot^2 24° = \cot^2 24° - \cot^2 24° = 0 = \) RHS
(vii) LHS \( = \sin^2 48° + \sin^2 42° = \sin^2(90° - 42°) + \sin^2 42° = \cos^2 42° + \sin^2 42° = 1 = \) RHS
(viii) LHS \( = \cos^2 57° - \sin^2 33° = \cos^2(90° - 33°) - \sin^2 33° = \sin^2 33° - \sin^2 33° = 0 = \) RHS
(ix) LHS \( = (\sin 65° + \cos 25°)(\sin 65° - \cos 25°) = \sin^2 65° - \cos^2 25° = \sin^2(90° - 25°) - \cos^2 25° = \cos^2 25° - \cos^2 25° = 0 = \) RHS
In simple words: Use complementary angle rules to rewrite angles so that the left and right sides of each equation become identical, then simplify to prove each statement.

Exam Tip: Always look for angle pairs that add up to 90° - these can be converted using complementary identities, often causing terms to cancel and leaving you with 0, 1, or another simple value.

Question 3. Prove the following.
(i) \( \sin 53° \cos 37° + \cos 53° \sin 37° = 1 \)
(ii) \( \cos 54° \cos 36° - \sin 54° \sin 36° = 0 \)
(iii) \( \sec 70° \sin 20° + \cos 20° \cosec 70° = 2 \)
Answer:
(i) LHS \( = \sin 53° \cos 37° + \cos 53° \sin 37° = \sin(90° - 37°) \cos 37° + \cos(90° - 37°) \sin 37° = \cos 37° \cos 37° + \sin 37° \sin 37° = \cos^2 37° + \sin^2 37° = 1 = \) RHS
(ii) LHS \( = \cos 54° \cos 36° - \sin 54° \sin 36° = \cos(90° - 36°) \cos 36° - \sin(90° - 36°) \sin 36° = \sin 36° \cos 36° - \cos 36° \sin 36° = 0 = \) RHS
(iii) LHS \( = \sec 70° \sin 20° + \cos 20° \cosec 70° = \sec(90° - 20°) \sin 20° + \cos 20° \cosec(90° - 20°) = \cosec 20° \cdot \frac{1}{\cosec 20°} + \frac{1}{\sec 20°} \cdot \sec 20° = 1 + 1 = 2 = \) RHS
In simple words: Convert angles using complementary identities, then use the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \) or reciprocal relationships to simplify and reach the required result.

Exam Tip: Watch for angle pairs that sum to 90° in mixed-angle expressions. Convert them consistently and look for products of reciprocals (like \( \cot \theta \times \tan \theta = 1 \)) to collapse the expression quickly.

Question 4. Prove the following.
(i) \( \frac{\sin 70°}{\cos 20°} + \frac{\cosec 20°}{\sec 70°} - 2\cos 70° \cosec 20° = 0 \)
(ii) \( \frac{\cos 80°}{\sin 10°} + \cos 59° \cosec 31° = 2 \)
(iii) \( \frac{2 \sin 68°}{\cos 22°} - \frac{2 \cot 15°}{5 \tan 75°} - \frac{3 \tan 45° \tan 20° \tan 40° \tan 50° \tan 70°}{5} = 1 \)
(iv) \( \frac{\sin 18°}{\cos 72°} + \sqrt{3}(\tan 10° \tan 30° \tan 40° \tan 50° \tan 80°) = 2 \)
(v) \( \frac{7 \cos 55°}{3 \sin 35°} - \frac{4(\cos 70° \cosec 20°)}{3(\tan 5° \tan 25° \tan 45° \tan 65° \tan 85°)} = 1 \)
Answer:
(i) LHS \( = \frac{\sin 70°}{\cos 20°} + \frac{\cosec 20°}{\sec 70°} - 2\cos 70° \cosec 20° = \frac{\sin 70°}{\sin(90° - 20°)} + \frac{\sec(90° - 20°)}{\sec 70°} - 2\cos 70° \sec(90° - 20°) = \frac{\sin 70°}{\sin 70°} + \frac{\sec 70°}{\sec 70°} - 2\cos 70° \sec 70° = 1 + 1 - 2 \times \cos 70° \times \frac{1}{\cos 70°} = 2 - 2 = 0 = \) RHS
(ii) LHS \( = \frac{\cos 80°}{\sin 10°} + \cos 59° \cosec 31° = \frac{\cos 80°}{\sin(90° - 10°)} + \sin(90° - 59°) \cosec 31° = \frac{\cos 80°}{\cos 80°} + \sin 31° \cosec 31° = 1 + \sin 31° \times \frac{1}{\sin 31°} = 1 + 1 = 2 = \) RHS
(iii) LHS \( = \frac{2 \sin 68°}{\sin(90° - 22°)} - \frac{2 \cot 15°}{5 \tan(90° - 75°)} - \frac{3 \times 1 \times \cot(90° - 20°) \times \cot(90° - 40°) \times \tan 50° \times \tan 70°}{5} = \frac{2 \sin 68°}{\sin 68°} - \frac{2 \cot 15°}{5 \cot 15°} - \frac{3 \times \cot 70° \cot 50° \tan 50° \tan 70°}{5} = 2 - \frac{2}{5} - \frac{3 \times \frac{1}{\tan 70°} \times \frac{1}{\tan 50°} \times \tan 50° \times \tan 70°}{5} = 2 - \frac{2}{5} - \frac{3}{5} = \frac{10 - 2 - 3}{5} = \frac{5}{5} = 1 = \) RHS
(iv) LHS \( = \frac{\sin 18°}{\sin(90° - 72°)} + \sqrt{3}[\cot(90° - 10°) \times \frac{1}{\sqrt{3}} \times \cot(90° - 40°) \times \tan 50° \times \tan 80°] = \frac{\sin 18°}{\sin 18°} + \sqrt{3} \times \frac{\cot 80° \times \cot 50° \times \tan 50° \times \tan 80°}{\sqrt{3}} = 1 + \frac{1}{\tan 80°} \times \frac{1}{\tan 50°} \times \tan 50° \times \tan 80° = 1 + 1 = 2 = \) RHS
(v) LHS \( = \frac{7 \cos 55°}{3 \cos(90° - 35°)} - \frac{4[\sin(90° - 70°) \cosec 20°]}{3[\cot(90° - 5°) \times \cot(90° - 25°) \times 1 \times \tan 65° \times \tan 85°]} = \frac{7 \cos 55°}{3 \cos 55°} - \frac{4(\sin 20° \cosec 20°)}{3(\cot 85° \cot 65° \tan 65° \tan 85°)} = \frac{7}{3} - \frac{4(\sin 20° \times \frac{1}{\sin 20°})}{3(\frac{1}{\tan 85°} \times \frac{1}{\tan 65°} \times \tan 65° \times \tan 85°)} = \frac{7}{3} - \frac{4}{3} = \frac{3}{3} = 1 = \) RHS
In simple words: Convert all angles to their complementary forms and apply reciprocal identities so that matched terms cancel, leaving simpler expressions that combine to the target value.

Exam Tip: In multi-step proofs involving many angles, first identify and group complementary angle pairs. Use the identity \( \tan \theta \cot \theta = 1 \) to eliminate products, and carefully track which terms cancel at each stage.

Question 5. Prove the following.
(i) \( \sin \theta \cos(90° - \theta) + \sin(90° - \theta) \cos \theta = 1 \)
(ii) \( \frac{\sin \theta}{\cos(90° - \theta)} + \frac{\cos \theta}{\sin(90° - \theta)} = 2 \)
(iii) \( \frac{\sin \theta \cos(90° - \theta) \cos \theta}{\sin(90° - \theta)} + \frac{\cos \theta \sin(90° - \theta) \sin \theta}{\cos(90° - \theta)} = 1 \)
(iv) \( \frac{\cos(90° - \theta) \sec(90° - \theta) \tan \theta}{\cosec(90° - \theta) \sin(90° - \theta) \cot(90° - \theta)} + \frac{\tan(90° - \theta)}{\cot \theta} = 2 \)
Answer:
(i) LHS \( = \sin \theta \cos(90° - \theta) + \sin(90° - \theta) \cos \theta = \sin \theta \sin \theta + \cos \theta \cos \theta = \sin^2 \theta + \cos^2 \theta = 1 = \) RHS. Hence proved.
(ii) LHS \( = \frac{\sin \theta}{\cos(90° - \theta)} + \frac{\cos \theta}{\sin(90° - \theta)} = \frac{\sin \theta}{\sin \theta} + \frac{\cos \theta}{\cos \theta} = 1 + 1 = 2 = \) RHS. Hence proved.
(iii) LHS \( = \frac{\sin \theta \cos(90° - \theta) \cos \theta}{\sin(90° - \theta)} + \frac{\cos \theta \sin(90° - \theta) \sin \theta}{\cos(90° - \theta)} = \frac{\sin \theta \sin \theta \cos \theta}{\cos \theta} + \frac{\cos \theta \cos \theta \sin \theta}{\sin \theta} = \sin^2 \theta + \cos^2 \theta = 1 = \) RHS. Hence proved.
(iv) LHS \( = \frac{\cos(90° - \theta) \sec(90° - \theta) \tan \theta}{\cosec(90° - \theta) \sin(90° - \theta) \cot(90° - \theta)} + \frac{\tan(90° - \theta)}{\cot \theta} = \frac{\sin \theta \cosec \theta \tan \theta}{\sec \theta \cos \theta \tan \theta} + \frac{\cot \theta}{\cot \theta} = 1 + 1 = 2 = \) RHS. Hence proved.
In simple words: Replace every complementary angle expression with its equivalent standard trig function using the 90° difference identities, then use basic ratios and the Pythagorean identity to simplify.

Exam Tip: Memorize the six complementary angle identities thoroughly. Once substituted, most of these problems reduce to simple cancellations or applications of \( \sin^2 \theta + \cos^2 \theta = 1 \).

Question 6. Prove the following.
(v) \( \frac{\cos(90° - \theta)}{1 + \sin(90° - \theta)} + \frac{1 + \sin(90° - \theta)}{\cos(90° - \theta)} = 2 \cosec \theta \)
(vi) \( \frac{\sec(90° - \theta) \cosec \theta - \tan(90° - \theta) \cot \theta + \cos^2 25° + \cos^2 65°}{3 \tan 27° \tan 63°} = \frac{2}{3} \)
(vii) \( \cot \theta \tan(90° - \theta) - \sec(90° - \theta) \cosec \theta + \sqrt{3} \tan 12° \tan 60° \tan 78° = 2 \)
Answer:
(v) LHS \( = \frac{\sin \theta}{1 + \cos \theta} + \frac{1 + \cos \theta}{\sin \theta} = \frac{\sin^2 \theta + (1 + \cos \theta)^2}{(1 + \cos \theta) \sin \theta} = \frac{\sin^2 \theta + 1 + \cos^2 \theta + 2 \cos \theta}{(1 + \cos \theta) \sin \theta} = \frac{1 + 1 + 2 \cos \theta}{(1 + \cos \theta) \sin \theta} = \frac{2 + 2 \cos \theta}{(1 + \cos \theta) \sin \theta} = \frac{2(1 + \cos \theta)}{(1 + \cos \theta) \sin \theta} = \frac{2}{\sin \theta} = 2 \cosec \theta = \) RHS. Hence proved.
(vi) LHS \( = \frac{\cosec \theta \cosec \theta - \cot \theta \cot \theta + \sin^2(90° - 25°) + \cos^2 65°}{3 \tan 27° \cot(90° - 63°)} = \frac{\cosec^2 \theta - \cot^2 \theta + \sin^2 65° + \cos^2 65°}{3 \tan 27° \cot 27°} = \frac{1 + 1}{3 \times \tan 27° \times \frac{1}{\tan 27°}} = \frac{2}{3} = \) RHS. Hence proved.
(vii) LHS \( = \cot \theta \cot \theta - \cosec \theta \cosec \theta + \sqrt{3} \tan 12° \times \sqrt{3} \times \cot(90° - 78°) = \cot^2 \theta - \cosec^2 \theta + 3 \tan 12° \cot 12° = -1 + 3 \times \tan 12° \times \frac{1}{\tan 12°} = -1 + 3 = 2 = \) RHS. Hence proved.
In simple words: Work systematically through complementary angle substitutions, combine fractions where needed, and apply Pythagorean identities and reciprocal cancellations to arrive at the target expression.

Exam Tip: For fractions, find a common denominator before substituting angles. For products of complementary angles with special values (like 12°, 27°, 60°), group them strategically so reciprocals cancel.

Question 7. Prove the following.
(i) \( \tan 5° \tan 25° \tan 30° \tan 65° \tan 85° = \frac{1}{\sqrt{3}} \)
(ii) \( \cot 12° \cot 38° \cot 52° \cot 60° \cot 78° = \frac{1}{\sqrt{3}} \)
(iii) \( \cos 15° \cos 35° \cosec 55° \cos 60° \cosec 75° = \frac{1}{2} \)
(iv) \( \cos 1° \cos 2° \cos 3° \ldots \cos 180° = 0 \)
(v) \( \left(\frac{\sin 49°}{\cos 41°}\right)^2 + \left(\frac{\cos 41°}{\sin 49°}\right)^2 = 2 \)
Answer:
(i) LHS \( = \tan(90° - 85°) \tan(90° - 65°) \times \frac{1}{\sqrt{3}} \times \frac{1}{\cot 60°} \times \frac{1}{\cot 85°} = \cot 85° \cot 65° \times \frac{1}{\sqrt{3}} \times \frac{1}{\cot 60°} \times \frac{1}{\cot 85°} = \frac{1}{\sqrt{3}} = \) RHS
(ii) LHS \( = \tan(90° - 12°) \times \tan(90° - 38°) \times \cot 52° \times \frac{1}{\sqrt{3}} \times \cot 78° = \frac{1}{\sqrt{3}} \times \tan 78° \times \tan 52° \times \cot 52° \times \cot 78° = \frac{1}{\sqrt{3}} \times \tan 78° \times \tan 52° \times \frac{1}{\tan 52°} \times \frac{1}{\tan 78°} = \frac{1}{\sqrt{3}} = \) RHS
(iii) LHS \( = \cos(90° - 75°) \cos(90° - 55°) \times \frac{1}{\sin 55°} \times \frac{1}{2} \times \frac{1}{\sin 75°} = \sin 75° \sin 55° \times \frac{1}{\sin 55°} \times \frac{1}{2} \times \frac{1}{\sin 75°} = \frac{1}{2} = \) RHS
(iv) LHS \( = \cos 1° \times \cos 2° \times \cos 3° \times \ldots \times \cos 90° \times \ldots \times \cos 180° = \cos 1° \times \cos 2° \times \cos 3° \times \ldots \times 0 \times \ldots \times \cos 180° = 0 = \) RHS
(v) LHS \( = \left(\frac{\cos(90° - 49°)}{\cos 41°}\right)^2 + \left(\frac{\cos 41°}{\cos(90° - 49°)}\right)^2 = \left(\frac{\cos 41°}{\cos 41°}\right)^2 + \left(\frac{\cos 41°}{\cos 41°}\right)^2 = 1^2 + 1^2 = 1 + 1 = 2 = \) RHS
In simple words: Pair complementary angles and use identities to convert them to matching forms. Products of reciprocals equal 1, and Pythagorean relationships simplify to the target value.

Exam Tip: When you see many angles in a product, look for complementary pairs (angles that sum to 90°). These often cancel or produce 1, dramatically simplifying the calculation. Watch for \( \cos 90° = 0 \), which will zero out any product containing it.

Question 8. Convert each of the following into an expression involving only acute angles.
(i) \( \sin 67° + \cos 75° \)
(ii) \( \cot 65° + \tan 49° \)
(iii) \( \sec 78° + \cosec 56° \)
(iv) \( \cosec 54° + \sin 72° \)
Answer:
(i) \( \sin 67° + \cos 75° = \cos(90° - 67°) + \sin(90° - 75°) = \cos 23° + \sin 15° \)
(ii) \( \cot 65° + \tan 49° = \cos(90° - 65°) + \cot(90° - 49°) = \cos 25° + \cot 41° \)
(iii) \( \sec 78° + \cosec 56° = \sec(90° - 12°) + \cosec(90° - 34°) = \cosec 12° + \sec 34° \)
(iv) \( \cosec 54° + \sin 72° = \sec(90° - 54°) + \cos(90° - 72°) = \sec 36° + \cos 18° \)
In simple words: Apply complementary angle formulas to each trig function. Rewrite angles larger than 45° as complements of smaller acute angles, which converts the original function to its cofunction.

Exam Tip: Always reduce angles to their acute-angle equivalents. If an angle is greater than 45°, it is easier to express it as the complement of a smaller angle, then swap the function accordingly.

Question 9. In triangle ABC, prove that \( \tan\left(\frac{C + A}{2}\right) = \cot\frac{B}{2} \).
Answer: In triangle ABC, we know that \( A + B + C = 180° \), so \( A + C = 180° - B \).

Now, LHS \( = \tan\left(\frac{C + A}{2}\right) = \tan\left(\frac{180° - B}{2}\right) = \tan\left(90° - \frac{B}{2}\right) = \cot\frac{B}{2} = \) RHS. Hence proved.
In simple words: Use the fact that the sum of angles in a triangle is 180°. Once you express \( C + A \) in terms of \( B \), the complementary angle identity converts the tangent to cotangent immediately.

Exam Tip: In triangle problems, always start with the angle-sum property. Converting half-angles often leads to complementary forms that simplify elegantly using the 90° identity.

Question 10. If \( \cos 2\theta = \sin 4\theta \), find the value of \( \theta \).
Answer: Starting with \( \cos 2\theta = \sin 4\theta \), we convert the cosine to a sine using the complementary angle identity:

\( \sin(90° - 2\theta) = \sin 4\theta \)

Comparing both sides: \( 90° - 2\theta = 4\theta \)

\( 90° = 6\theta \)

\( \theta = \frac{90°}{6} = 15° \)

Hence, the value of \( \theta \) is \( 15° \).
In simple words: Convert the cosine term using the complementary angle rule, then equate the angles inside the sine functions. Solve the resulting linear equation for \( \theta \).

Exam Tip: Always convert all trigonometric functions to the same type (usually sine or cosine) before equating arguments. This eliminates confusion and makes the algebraic solution straightforward.

Question 11. If \( \sec 2A = \cosec(A - 42°) \), find the value of \( A \).
Answer: Starting with \( \sec 2A = \cosec(A - 42°) \), we convert the secant using the complementary angle identity:

\( \cosec(90° - 2A) = \cosec(A - 42°) \)

Comparing both sides: \( 90° - 2A = A - 42° \)

\( 90° + 42° = A + 2A \)

\( 3A = 132° \)

\( A = \frac{132°}{3} = 44° \)

Hence, the value of \( A \) is \( 44° \).
In simple words: Use the identity \( \sec \theta = \cosec(90° - \theta) \) to rewrite the secant. Then match the arguments of the two cosecant functions and solve for \( A \).

Exam Tip: Keep the complementary angle formula sheet in mind - \( \sec \theta = \cosec(90° - \theta) \) and \( \tan \theta = \cot(90° - \theta) \) are frequently tested. After conversion, simple equation-solving follows.

Question 12. If \( \sin 3A = \cos(A - 26°) \), find the value of \( A \).
Answer: Starting with \( \sin 3A = \cos(A - 26°) \), we convert the sine using the complementary angle identity:

\( \cos(90° - 3A) = \cos(A - 26°) \)

Comparing both sides: \( 90° - 3A = A - 26° \)

\( 90° + 26° = A + 3A \)

\( 4A = 116° \)

\( A = \frac{116°}{4} = 29° \)

Hence, the value of \( A \) is \( 29° \).
In simple words: Convert sine to cosine using \( \sin \theta = \cos(90° - \theta) \). Equate the arguments of the two cosine expressions and solve the resulting linear equation for \( A \).

Exam Tip: When comparing \( \cos \alpha = \cos \beta \), equate the angles directly (i.e., \( \alpha = \beta \)). This works because cosine is a one-to-one function within the standard angle range.