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Trigonometric Identities and Exercises

🎓 Class 10 Mathematics CBSE Theory Ch 8 — Introduction to Trigonometry ⏱ ~35 min
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This MCQ module is based on: Trigonometric Identities and Exercises

This mathematics assessment will be based on: Trigonometric Identities and Exercises
Targeting Class 10 level in Trigonometry, with Intermediate difficulty.

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8.4 Trigonometric Identities

An equation involving trigonometric ratios of an angle is called a trigonometric identityi if it is true for all values of the angle for which both sides are defined.

A B C Opp Adj Hyp
Fig. 8.21 — Right triangle used for identity proofs

By Pythagoras: \(AB^2+BC^2=AC^2\). Divide throughout by \(AC^2\):

\[\left(\frac{AB}{AC}\right)^2+\left(\frac{BC}{AC}\right)^2=1\Rightarrow \boxed{\;\cos^2 A+\sin^2 A=1\;}\]

Divide instead by \(AB^2\):

\[1+\left(\frac{BC}{AB}\right)^2=\left(\frac{AC}{AB}\right)^2\Rightarrow \boxed{\;1+\tan^2 A=\sec^2 A\;}\]

Divide by \(BC^2\):

\[\left(\frac{AB}{BC}\right)^2+1=\left(\frac{AC}{BC}\right)^2\Rightarrow \boxed{\;\cot^2 A+1=\csc^2 A\;}\]
Three Fundamental Identities
For \(0°\le A\le 90°\) (with appropriate restrictions where the ratios are undefined):
  1. \(\sin^2 A+\cos^2 A=1\)
  2. \(\sec^2 A-\tan^2 A=1\) for \(0°\le A<90°\)
  3. \(\csc^2 A-\cot^2 A=1\) for \(0°

Using these, if any one trigonometric ratio of A is known, all others can be derived. For instance, if \(\tan A=1/\sqrt3\), then \(\sec^2 A=1+1/3=4/3\Rightarrow \sec A=2/\sqrt3\Rightarrow \cos A=\sqrt3/2\), then \(\sin A=1/2\).

Example 9

Express \(\cos A\), \(\tan A\) and \(\sec A\) in terms of \(\sin A\).

Solution. \(\cos^2 A=1-\sin^2 A\Rightarrow \cos A=\sqrt{1-\sin^2 A}\). Then \(\tan A=\sin A/\cos A=\dfrac{\sin A}{\sqrt{1-\sin^2 A}}\) and \(\sec A=\dfrac{1}{\sqrt{1-\sin^2 A}}\).

Example 10

Prove: \(\sec A(1-\sin A)(\sec A+\tan A)=1\).

Solution. LHS \(=\dfrac{1}{\cos A}(1-\sin A)\left(\dfrac{1}{\cos A}+\dfrac{\sin A}{\cos A}\right)=\dfrac{(1-\sin A)(1+\sin A)}{\cos^2 A}=\dfrac{1-\sin^2 A}{\cos^2 A}=\dfrac{\cos^2 A}{\cos^2 A}=1.\)

Example 11

Prove: \(\dfrac{\cot A-\cos A}{\cot A+\cos A}=\dfrac{\csc A-1}{\csc A+1}\).

Solution. LHS \(=\dfrac{\frac{\cos A}{\sin A}-\cos A}{\frac{\cos A}{\sin A}+\cos A}=\dfrac{\cos A(1/\sin A-1)}{\cos A(1/\sin A+1)}=\dfrac{\csc A-1}{\csc A+1}=\) RHS.

Activity — Identity Verification Grid

Pick three angles (say 30°, 45°, 60°) and verify numerically that \(\sin^2\theta+\cos^2\theta=1\) in each case. Record your calculations.

30°: 1/4 + 3/4 = 1 ✓. 45°: 1/2 + 1/2 = 1 ✓. 60°: 3/4 + 1/4 = 1 ✓.

Exercise 8.1

Q1. In \(\triangle ABC\), right-angled at B, AB = 24 cm, BC = 7 cm. Determine (i) sin A, cos A; (ii) sin C, cos C.
AC = √(576+49) = 25. (i) sin A = BC/AC = 7/25, cos A = AB/AC = 24/25. (ii) sin C = AB/AC = 24/25, cos C = BC/AC = 7/25.
Q2. In Fig 8.13, find tan P − cot R.
With PQ = 12, PR = 13 ⇒ QR = 5. tan P = QR/PQ = 5/12. cot R = QR/PQ = 5/12. Difference = 0.
Q3. If sin A = 3/4, calculate cos A and tan A.
cos A = √(1−9/16) = √7/4. tan A = sin/cos = 3/√7.
Q4. Given 15 cot A = 8, find sin A and sec A.
cot A = 8/15 ⇒ tan A = 15/8 ⇒ sides opp:adj = 15:8, hyp = 17. sin A = 15/17, sec A = 17/8.
Q5. Given sec θ = 13/12, compute all other trigonometric ratios.
cos θ = 12/13, sin θ = √(1 − 144/169) = 5/13. tan = 5/12, cot = 12/5, cosec = 13/5.
Q6. If ∠A and ∠B are acute such that cos A = cos B, show that ∠A = ∠B.
In right triangles sharing this property, adjacent/hypotenuse is equal; triangles are similar ⇒ A = B.
Q7. If cot θ = 7/8, evaluate (i) (1+sin θ)(1−sin θ)/(1+cos θ)(1−cos θ); (ii) cot²θ.
(i) = cos²θ/sin²θ = cot²θ = 49/64. (ii) 49/64.
Q8. If 3 cot A = 4, check whether (1 − tan²A)/(1 + tan²A) = cos²A − sin²A or not.
cot A = 4/3, tan A = 3/4. LHS = (1 − 9/16)/(1 + 9/16) = (7/16)/(25/16) = 7/25. cos²A − sin²A with cos = 4/5, sin = 3/5: 16/25 − 9/25 = 7/25. Equal. ✓
Q9. In \(\triangle ABC\) right-angled at B, if tan A = 1/√3, find (i) sin A cos C + cos A sin C, (ii) cos A cos C − sin A sin C.
A = 30°, C = 60°. (i) = sin(A+C) = sin 90° = 1. (ii) = cos(A+C) = cos 90° = 0.
Q10. In \(\triangle PQR\) right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Find sin P, cos P, tan P.
Let QR = x, PR = 25 − x. PR² = PQ² + QR²: (25−x)² = 25 + x² ⇒ 625 − 50x = 25 ⇒ x = 12. So QR = 12, PR = 13. sin P = 12/13, cos P = 5/13, tan P = 12/5.
Q11. State whether true or false: (i) tan A is always less than 1; (ii) sec A = 12/5 for some A; (iii) cos A is the abbreviation for cosecant A; (iv) cot A is the product of cot and A; (v) sin θ = 4/3 for some θ.
(i) False — e.g. tan 60° = √3 > 1. (ii) True — since 12/5 > 1, valid secant value. (iii) False — cos is cosine, cosec is cosecant. (iv) False — cot A is a single symbol. (v) False — sin of any angle is ≤ 1.

Exercise 8.2

Q1. Evaluate: (i) sin 60° cos 30° + sin 30° cos 60°; (ii) 2 tan²45° + cos²30° − sin²60°; (iii) cos 45° / (sec 30° + cosec 30°); (iv) (sin 30° + tan 45° − cosec 60°)/(sec 30° + cos 60° + cot 45°); (v) (5 cos²60° + 4 sec²30° − tan²45°)/(sin²30° + cos²30°).
(i) (√3/2)(√3/2) + (1/2)(1/2) = 1. (ii) 2 + 3/4 − 3/4 = 2. (iii) (1/√2)/(2/√3 + 2) = (√3(√3 − 1))/(4·√2) after rationalising = (3√2 − √6)/8. (iv) After simplification = (4√3 − 3√3)/(… ) — common answer (43 − 24√3)/11. (v) 5·1/4 + 4·4/3 − 1 = 5/4 + 16/3 − 1 = (15 + 64 − 12)/12 = 67/12. Denominator = 1. Answer 67/12.
Q2. Choose correct option (i) 2tan30°/(1+tan²30°).
2(1/√3)/(1 + 1/3) = (2/√3)/(4/3) = 6/(4√3) = √3/2 = sin 60°. Option (D).
Q3. If tan(A+B) = √3 and tan(A−B) = 1/√3, 0° < A+B ≤ 90°, A > B, find A and B.
A + B = 60°, A − B = 30°. Solving: A = 45°, B = 15°.

Exercise 8.3

Q1. Express sin A, sec A and tan A in terms of cot A.
tan A = 1/cot A. sec A = √(1+tan²A) = √(1 + 1/cot²A) = √(cot²A + 1)/cot A. sin A = 1/√(1+cot²A).
Q2. Write all other trigonometric ratios of ∠A in terms of sec A.
cos A = 1/sec A. tan A = √(sec²A − 1). sin A = √(sec²A − 1)/sec A. cot A = 1/√(sec²A − 1). cosec A = sec A /√(sec²A − 1).
Q3. Choose the correct option. (i) 9 sec²A − 9 tan²A = ?
= 9(sec²A − tan²A) = 9·1 = 9. (B).
Q4. Prove (cosec θ − cot θ)² = (1 − cos θ)/(1 + cos θ).
LHS = ((1 − cos θ)/sin θ)² = (1 − cos θ)²/sin²θ = (1 − cos θ)²/(1 − cos²θ) = (1 − cos θ)²/((1 − cos θ)(1 + cos θ)) = (1 − cos θ)/(1 + cos θ) = RHS.
Q5. Prove cos A/(1 + sin A) + (1 + sin A)/cos A = 2 sec A.
LHS = [cos²A + (1 + sin A)²] / [(1 + sin A)·cos A] = [cos²A + 1 + 2 sin A + sin²A]/… = [2 + 2 sin A]/[(1 + sin A)cos A] = 2(1 + sin A)/[(1 + sin A)cos A] = 2/cos A = 2 sec A.
Q6. Prove tan θ/(1 − cot θ) + cot θ/(1 − tan θ) = 1 + sec θ cosec θ.
Writing everything in sin, cos and simplifying, the LHS reduces to (sin³θ − cos³θ)/(sin θ cos θ (sin θ − cos θ)) = (sin²θ + sin θ cos θ + cos²θ)/(sin θ cos θ) = 1 + 1/(sin θ cos θ) = 1 + sec θ · cosec θ.
Competency-Based Questions
Q1. If sin θ + cos θ = √2, find sin θ cos θ.
L3 Apply
Squaring: 1 + 2 sin θ cos θ = 2 ⇒ sin θ cos θ = 1/2. (Occurs when θ = 45°.)
Q2. A surveyor writes the identity "sec²A − tan²A = 0". Diagnose the error.
L4 Analyse
Wrong. The correct identity is sec²A − tan²A = 1. The surveyor may have confused it with sin²A − sin²A = 0. Always verify at A = 0°: sec²0 − tan²0 = 1 − 0 = 1 ✓.
Q3. Evaluate whether √((1 + sin A)/(1 − sin A)) = sec A + tan A for every acute A.
L5 Evaluate
(1 + sin A)/(1 − sin A) multiplied numerator & denominator by (1 + sin A): (1 + sin A)²/(1 − sin²A) = (1 + sin A)²/cos²A. √ gives (1 + sin A)/cos A = sec A + tan A. Identity is true for 0° ≤ A < 90°.
Q4. Create an original identity in terms of tan θ by starting with sec²θ − tan²θ = 1 and multiplying both sides by sin²θ.
L6 Create
sin²θ sec²θ − sin²θ tan²θ = sin²θ ⇒ tan²θ − sin²θ tan²θ = sin²θ ⇒ tan²θ (1 − sin²θ) = sin²θ ⇒ tan²θ cos²θ = sin²θ, which is equivalent to the definition tan θ = sin θ / cos θ. A valid derived identity.
Assertion & Reason
A: 1 + tan²A = sec²A for every acute A.
R: This follows from dividing the Pythagorean relation AB² + BC² = AC² by AB².
A) Both true; R explains A
B) Both true; R does not explain A
C) A true, R false
D) A false, R true
Answer: A. Dividing by AB² gives 1 + (BC/AB)² = (AC/AB)², i.e. 1 + tan²A = sec²A.
A: cos θ = √(1 − sin²θ) is valid for acute θ only.
R: For acute θ, cos θ > 0 so the positive square root is taken.
A) Both true; R explains A
B) Both true; R does not explain A
C) A true, R false
D) A false, R true
Answer: A. For obtuse angles cos θ can be negative, so we'd need −√(…). The positive-root form works precisely because cos θ > 0 for acute θ.
A: cot A is not defined at A = 0°.
R: cot A = cos A / sin A, and sin 0° = 0.
A) Both true; R explains A
B) Both true; R does not explain A
C) A true, R false
D) A false, R true
Answer: A. Division by sin 0° = 0 is undefined, so cot 0° has no value.

8.5 Summary

Key Results
  1. In a right triangle, the trigonometric ratios of an acute angle A are: \(\sin A=\frac{\text{opp}}{\text{hyp}},\cos A=\frac{\text{adj}}{\text{hyp}},\tan A=\frac{\text{opp}}{\text{adj}}\); and their reciprocals \(\csc A,\sec A,\cot A\).
  2. \(\tan A=\dfrac{\sin A}{\cos A}\), \(\cot A=\dfrac{\cos A}{\sin A}\).
  3. Knowing any one ratio determines all the others (up to the choice of the acute angle).
  4. Standard values are tabulated for 0°, 30°, 45°, 60° and 90°.
  5. \(\sin A\) increases, \(\cos A\) decreases as \(A\) increases from 0° to 90°; both always lie between 0 and 1.
  6. Identities: \(\sin^2 A+\cos^2 A=1\); \(\sec^2 A-\tan^2 A=1\) for \(0°\le A<90°\); \(\csc^2 A-\cot^2 A=1\) for \(0°
Looking Ahead
In the next chapter — "Some Applications of Trigonometry" — you will use these ratios together with the ideas of angle of elevation and angle of depression to solve real-world problems involving heights and distances.

Frequently Asked Questions

How is the identity sin squared A plus cos squared A = 1 proved?
In a right triangle with hypotenuse c, opposite a and adjacent b, by Pythagoras a squared plus b squared = c squared. Dividing by c squared gives (a/c) squared plus (b/c) squared = 1, which is sin squared A plus cos squared A = 1.
How do you prove an identity like (1 - cos A)/(1 + cos A) = (cosec A - cot A) squared?
Start from one side, rewrite using sin A and cos A, simplify using sin squared A = 1 - cos squared A, and manipulate algebraically until you reach the other side.
Why are trigonometric identities important?
They simplify expressions, help solve equations, prove other results and are fundamental in calculus, physics and engineering.
What is the summary of Chapter 8 Trigonometry?
Trigonometric ratios relate angle and side lengths in right triangles; specific angles (0, 30, 45, 60, 90 degrees) have standard exact values; and three fundamental identities derived from Pythagoras link the ratios.
How do exercises in Chapter 8 test identities?
Exercises include direct computation with specific angles, simplification of expressions, and proving identities by rewriting both sides using sin and cos and simplifying.
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