CBSE Class 11 Mathematics Trigonometric Functions Notes Set 09

Important Notes and Formulas

The Meaning of Trigonometry

Tri (3) + Gon (sides) + Metron (measure). Hence, this particular branch in Mathematics was developed in ancient past to measure 3 sides, 3 angles and 6 elements of a triangle. In today's time trigonometric functions are used in entirely different shapes. The 2 basic functions are sine and cosine of an angle in a right-angled triangle and there are 4 other derived functions.

In the right-angled triangle above, with P = perpendicular, B = base and H = hypotenuse:
sin θ = \( \frac{P}{H} \),   cos θ = \( \frac{B}{H} \),   tan θ = \( \frac{P}{B} \),   cot θ = \( \frac{B}{P} \),   sec θ = \( \frac{H}{B} \),   cosec θ = \( \frac{H}{P} \)

Basic Trigonometric Identities

(a) sin²θ + cos²θ = 1; −1 ≤ sin θ ≤ 1; −1 ≤ cos θ ≤ 1 for all θ ∈ R
(b) sec²θ − tan²θ = 1; |sec θ| ≥ 1 for all θ ∈ R (where defined)
(c) cosec²θ − cot²θ = 1; |cosec θ| ≥ 1 for all θ ∈ R (where defined)

Trigonometric Ratios of Standard Angles

Signs of Trigonometric Ratios in Different Quadrants

The sign of the trigonometric ratios in different quadrants are as under:

Trigonometric Ratios of Allied Angles

Using trigonometric ratios of allied angles, we can find the trigonometric ratios of angles of any magnitude.
\( \sin(-\theta) = -\sin\theta \),   \( \cos(-\theta) = \cos\theta \)
\( \tan(-\theta) = -\tan\theta \),   \( \cot(-\theta) = -\cot\theta \)
\( \text{cosec}(-\theta) = -\text{cosec}\,\theta \),   \( \sec(-\theta) = \sec\theta \)
\( \sin\left(\frac{\pi}{2} - \theta\right) = \cos\theta \),   \( \cos\left(\frac{\pi}{2} - \theta\right) = \sin\theta \)
\( \tan\left(\frac{\pi}{2} - \theta\right) = \cot\theta \),   \( \cot\left(\frac{\pi}{2} - \theta\right) = \tan\theta \)
\( \sec\left(\frac{\pi}{2} - \theta\right) = \text{cosec}\,\theta \),   \( \text{cosec}\left(\frac{\pi}{2} - \theta\right) = \sec\theta \)
\( \sin\left(\frac{\pi}{2} + \theta\right) = \cos\theta \),   \( \cos\left(\frac{\pi}{2} + \theta\right) = -\sin\theta \)
\( \tan\left(\frac{\pi}{2} + \theta\right) = -\cot\theta \),   \( \cot\left(\frac{\pi}{2} + \theta\right) = -\tan\theta \)
\( \sec\left(\frac{\pi}{2} + \theta\right) = -\text{cosec}\,\theta \),   \( \text{cosec}\left(\frac{\pi}{2} + \theta\right) = \sec\theta \)
\( \sin(\pi - \theta) = \sin\theta \),   \( \cos(\pi - \theta) = -\cos\theta \)
\( \tan(\pi - \theta) = -\tan\theta \),   \( \cot(\pi - \theta) = -\cot\theta \)
\( \sec(\pi - \theta) = -\sec\theta \),   \( \text{cosec}(\pi - \theta) = \text{cosec}\,\theta \)
\( \sin(\pi + \theta) = -\sin\theta \),   \( \cos(\pi + \theta) = -\cos\theta \)
\( \tan(\pi + \theta) = \tan\theta \),   \( \cot(\pi + \theta) = \cot\theta \)
\( \sec(\pi + \theta) = -\sec\theta \),   \( \text{cosec}(\pi + \theta) = -\text{cosec}\,\theta \)
\( \sin\left(\frac{3\pi}{2} - \theta\right) = -\cos\theta \),   \( \cos\left(\frac{3\pi}{2} - \theta\right) = -\sin\theta \)
\( \tan\left(\frac{3\pi}{2} - \theta\right) = \cot\theta \),   \( \cot\left(\frac{3\pi}{2} - \theta\right) = \tan\theta \)
\( \sec\left(\frac{3\pi}{2} - \theta\right) = -\text{cosec}\,\theta \),   \( \text{cosec}\left(\frac{3\pi}{2} - \theta\right) = -\sec\theta \)
\( \sin\left(\frac{3\pi}{2} + \theta\right) = -\cos\theta \),   \( \cos\left(\frac{3\pi}{2} + \theta\right) = \sin\theta \)
\( \tan\left(\frac{3\pi}{2} + \theta\right) = -\cot\theta \),   \( \cot\left(\frac{3\pi}{2} + \theta\right) = -\tan\theta \)
\( \sec\left(\frac{3\pi}{2} + \theta\right) = \text{cosec}\,\theta \),   \( \text{cosec}\left(\frac{3\pi}{2} + \theta\right) = -\sec\theta \)
\( \sin(2\pi - \theta) = -\sin\theta \),   \( \cos(2\pi - \theta) = \cos\theta \)
\( \tan(2\pi - \theta) = -\tan\theta \),   \( \cot(2\pi - \theta) = -\cot\theta \)
\( \sec(2\pi - \theta) = \sec\theta \),   \( \text{cosec}(2\pi - \theta) = -\text{cosec}\,\theta \)
\( \sin(2\pi + \theta) = \sin\theta \),   \( \cos(2\pi + \theta) = \cos\theta \)
\( \tan(2\pi + \theta) = \tan\theta \),   \( \cot(2\pi + \theta) = \cot\theta \)
\( \sec(2\pi + \theta) = \sec\theta \),   \( \text{cosec}(2\pi + \theta) = \text{cosec}\,\theta \)

Trigonometric Functions of Sum or Difference of Two Angles

(a) sin (A + B) = sin A cos B + cos A sin B
(b) sin (A − B) = sin A cos B − cos A sin B
(c) cos (A + B) = cos A cos B − sin A sin B
(d) cos (A − B) = cos A cos B + sin A sin B
(e) \( \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \)
(f) \( \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \)
(g) \( \cot(A + B) = \frac{\cot A \cot B - 1}{\cot B + \cot A} \)
(h) \( \cot(A - B) = \frac{\cot A \cot B + 1}{\cot B - \cot A} \)
(i) sin²A − sin²B = cos²B − cos²A = sin (A + B) · sin (A − B)
(j) cos²A − sin²B = cos²B − sin²A = cos (A + B) · cos (A − B)
(k) \( \tan(A + B + C) = \frac{\tan A + \tan B + \tan C - \tan A \tan B \tan C}{1 - \tan A \tan B - \tan B \tan C - \tan C \tan A} \)

Multiple Angles and Half Angles

(a) sin 2A = 2 sin A cos A;   \( \sin\theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2} \)
(b) cos 2A = cos²A − sin²A = 2 cos²A − 1 = 1 − 2 sin²A;   \( 2\cos^2\frac{\theta}{2} = 1 + \cos\theta \), \( 2\sin^2\frac{\theta}{2} = 1 - \cos\theta \)
(c) \( \tan 2A = \frac{2\tan A}{1 - \tan^2 A} \);   \( \tan\theta = \frac{2\tan\frac{\theta}{2}}{1 - \tan^2\frac{\theta}{2}} \)
(d) \( \sin 2A = \frac{2\tan A}{1 + \tan^2 A} \);   \( \cos 2A = \frac{1 - \tan^2 A}{1 + \tan^2 A} \)
(e) sin 3A = 3 sin A − 4 sin³A
(f) cos 3A = 4 cos³A − 3 cos A
(g) \( \tan 3A = \frac{3\tan A - \tan^3 A}{1 - 3\tan^2 A} \)

Transformation of Products into Sum or Difference of Sines and Cosines

(a) 2 sin A cos B = sin (A + B) + sin (A − B)
(b) 2 cos A sin B = sin (A + B) − sin (A − B)
(c) 2 cos A cos B = cos (A + B) + cos (A − B)
(d) 2 sin A sin B = cos (A − B) − cos (A + B)

Factorisation of the Sum or Difference of Two Sines or Cosines

(a) \( \sin C + \sin D = 2\sin\frac{C + D}{2}\cos\frac{C - D}{2} \)
(b) \( \sin C - \sin D = 2\cos\frac{C + D}{2}\sin\frac{C - D}{2} \)
(c) \( \cos C + \cos D = 2\cos\frac{C + D}{2}\cos\frac{C - D}{2} \)
(d) \( \cos C - \cos D = -2\sin\frac{C + D}{2}\sin\frac{C - D}{2} \)

Important Trigonometric Ratios

(a) sin nπ = 0;  cos nπ = (−1)ⁿ;  tan nπ = 0, where n ∈ Z
(b) \( \sin 15^\circ = \sin\frac{\pi}{12} = \frac{\sqrt{3} - 1}{2\sqrt{2}} = \cos 75^\circ = \cos\frac{5\pi}{12} \)
\( \cos 15^\circ = \cos\frac{\pi}{12} = \frac{\sqrt{3} + 1}{2\sqrt{2}} = \sin 75^\circ = \sin\frac{5\pi}{12} \)
\( \tan 15^\circ = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = 2 - \sqrt{3} = \cot 75^\circ \)
\( \tan 75^\circ = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} = 2 + \sqrt{3} = \cot 15^\circ \)
(c) \( \sin\frac{\pi}{10} = \sin 18^\circ = \frac{\sqrt{5} - 1}{4} \) and \( \cos 36^\circ = \cos\frac{\pi}{5} = \frac{\sqrt{5} + 1}{4} \)

Conditional Identities

If A + B + C = π, then:
(i) sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C
(ii) \( \sin A + \sin B + \sin C = 4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2} \)
(iii) cos 2A + cos 2B + cos 2C = −1 − 4 cos A cos B cos C
(iv) \( \cos A + \cos B + \cos C = 1 + 4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2} \)
(v) tan A + tan B + tan C = tan A tan B tan C
(vi) \( \tan\frac{A}{2}\tan\frac{B}{2} + \tan\frac{B}{2}\tan\frac{C}{2} + \tan\frac{C}{2}\tan\frac{A}{2} = 1 \)
(vii) \( \cot\frac{A}{2} + \cot\frac{B}{2} + \cot\frac{C}{2} = \cot\frac{A}{2}\cot\frac{B}{2}\cot\frac{C}{2} \)
(viii) cot A cot B + cot B cot C + cot C cot A = 1

Range of a Trigonometric Expression

For E = a sin θ + b cos θ:
\( E = \sqrt{a^2 + b^2}\,\sin(\theta + \alpha) \), where \( \tan\alpha = \frac{b}{a} \)
\( E = \sqrt{a^2 + b^2}\,\cos(\theta - \beta) \), where \( \tan\beta = \frac{a}{b} \)
Hence, for any real value of θ, \( -\sqrt{a^2 + b^2} \le E \le \sqrt{a^2 + b^2} \).

Graphs of Trigonometric Functions

(a) y = sin x, x ∈ R; y ∈ [−1, 1]

(b) y = cos x, x ∈ R; y ∈ [−1, 1]

(c) y = tan x, \( x \in R - \left\{(2n + 1)\frac{\pi}{2} : n \in Z\right\} \); y ∈ R

(d) y = cot x, x ∈ R − {nπ : n ∈ Z}; y ∈ R

(e) y = cosec x, x ∈ R − {nπ : n ∈ Z}; y ∈ (−∞, −1] ∪ [1, ∞)

(f) y = sec x, \( x \in R - \left\{(2n + 1)\frac{\pi}{2} : n \in Z\right\} \); y ∈ (−∞, −1] ∪ [1, ∞)

Solved Examples

Example 1. If sec α and cosec α are the roots of x² − px + q = 0, then show that p² = q (q + 2).
Answer:
Since sec α and cosec α are the roots of x² − px + q = 0,
∴ sec α + cosec α = p and sec α · cosec α = q
∴ sin α + cos α = p sin α cos α and \( \sin\alpha\cos\alpha = \frac{1}{q} \)
∴ \( \sin\alpha + \cos\alpha = \frac{p}{q} \)
Squaring both sides, we get \( \sin^2\alpha + \cos^2\alpha + 2\sin\alpha\cos\alpha = \frac{p^2}{q^2} \)
⇒ \( 1 + \frac{2}{q} = \frac{p^2}{q^2} \) ⇒ p² = q (q + 2).

Example 2. If α, β and γ are in A.P., show that \( \cot\beta = \frac{\sin\alpha - \sin\gamma}{\cos\gamma - \cos\alpha} \).
Answer:
Since α, β and γ are in A.P., 2β = α + γ
⇒ \( \cot\beta = \cot\left(\frac{\alpha + \gamma}{2}\right) = \frac{\cos\left(\frac{\alpha + \gamma}{2}\right)}{\sin\left(\frac{\alpha + \gamma}{2}\right)} \)
Multiplying numerator and denominator by \( 2\sin\left(\frac{\alpha - \gamma}{2}\right) \), we get
\( \cot\beta = \frac{2\cos\left(\frac{\alpha + \gamma}{2}\right)\sin\left(\frac{\alpha - \gamma}{2}\right)}{2\sin\left(\frac{\alpha + \gamma}{2}\right)\sin\left(\frac{\alpha - \gamma}{2}\right)} = \frac{\sin\alpha - \sin\gamma}{\cos\gamma - \cos\alpha} \)

Example 3. Prove that tan A + 2 tan 2A + 4 tan 4A + 8 cot 8A = cot A.
Answer:
L.H.S. \( = \tan A + 2\tan 2A + 4\tan 4A + 8\left(\frac{1 - \tan^2 4A}{2\tan 4A}\right) \)
\( = \tan A + 2\tan 2A + \frac{4\tan^2 4A + 4 - 4\tan^2 4A}{\tan 4A} \) = tan A + 2 tan 2A + 4 cot 4A
\( = \tan A + 2\tan 2A + 4\left(\frac{1 - \tan^2 2A}{2\tan 2A}\right) = \tan A + \frac{2\tan^2 2A + 2 - 2\tan^2 2A}{\tan 2A} \) = tan A + 2 cot 2A
\( = \tan A + 2\left(\frac{1 - \tan^2 A}{2\tan A}\right) = \frac{\tan^2 A + 1 - \tan^2 A}{\tan A} \) = cot A = R.H.S.
Note: Students are advised to learn the above result as a formula, i.e., tan A + 2 cot 2A = cot A.

Example 4. Prove that: tan A + tan (60° + A) − tan (60° − A) = 3 tan 3A
Answer:
LHS = tan A + tan (60° + A) − tan (60° − A)
\( = \tan A + \frac{\sqrt{3} + \tan A}{1 - \sqrt{3}\tan A} - \frac{\sqrt{3} - \tan A}{1 + \sqrt{3}\tan A} \)
\( = \tan A + \frac{8\tan A}{1 - 3\tan^2 A} = \frac{9\tan A - 3\tan^3 A}{1 - 3\tan^2 A} \)
\( = 3\left(\frac{3\tan A - \tan^3 A}{1 - 3\tan^2 A}\right) \) = 3 tan 3A = RHS

Example 5. Prove that: \( \cos^3 A + \cos^3(120^\circ + A) + \cos^3(240^\circ + A) = \frac{3}{4}\cos 3A \)
Answer:
We know that cos 3A = 4 cos³A − 3 cos A ⇒ \( \cos^3 A = \frac{1}{4}(\cos 3A + 3\cos A) \)
∴ LHS \( = \frac{1}{4}\{\cos 3A + 3\cos A\} + \frac{1}{4}\{\cos(360^\circ + 3A) + 3\cos(120^\circ + A)\} + \frac{1}{4}\{\cos(720^\circ + 3A) + 3\cos(240^\circ + A)\} \)
\( = \frac{3}{4}\cos 3A + \frac{3}{4}\{\cos A + \cos(120^\circ + A) + \cos(240^\circ + A)\} \)
\( = \frac{3}{4}\cos 3A + \frac{3}{4}\{\cos A + 2\cos(180^\circ + A)\cos 60^\circ\} = \frac{3}{4}\cos 3A + \frac{3}{4}\left\{\cos A - 2\cos A \times \frac{1}{2}\right\} = \frac{3}{4}\cos 3A \) = RHS
Alternative method: cos A + cos (120° + A) + cos (240° + A) = cos A + 2 cos (180° + A) cos 60° = cos A − cos A = 0
∴ cos³A + cos³(120° + A) + cos³(240° + A) = 3 cos A cos (120° + A) cos (240° + A)  [∵ a + b + c = 0 ⇒ a³ + b³ + c³ = 3abc]
= 3 cos A cos (180° − (60° − A)) cos (180° + (60° + A)) = 3 cos A cos (60° − A) cos (60° + A)
\( = 3 \times \frac{1}{4}\cos 3A = \frac{3}{4}\cos 3A \)  [∵ \( \cos A\cos(60^\circ - A)\cos(60^\circ + A) = \frac{1}{4}\cos 3A \)]

Example 6. Prove that: sin 3A sin³A + cos 3A cos³A = cos³2A
Answer:
We have \( \cos^3 A = \frac{\cos 3A + 3\cos A}{4} \) and \( \sin^3 A = \frac{3\sin A - \sin 3A}{4} \)
∴ LHS \( = \sin 3A\left\{\frac{3\sin A - \sin 3A}{4}\right\} + \cos 3A\left\{\frac{\cos 3A + 3\cos A}{4}\right\} \)
\( = \frac{1}{4}\{3(\cos A\cos 3A + \sin A\sin 3A) + (\cos^2 3A - \sin^2 3A)\} = \frac{1}{4}\{3\cos(3A - A) + \cos 2(3A)\} \)
\( = \frac{1}{4}\{3\cos 2A + \cos 3(2A)\} = \frac{1}{4}\{3\cos 2A + (4\cos^3 2A - 3\cos 2A)\} \) = cos³2A = RHS

Example 7. Prove that: tan 6° tan 42° tan 66° tan 78° = 1
Answer:
\( \text{LHS} = \frac{\sin 6^\circ\sin 42^\circ\sin 66^\circ\sin 78^\circ}{\cos 6^\circ\cos 42^\circ\cos 66^\circ\cos 78^\circ} = \frac{(2\sin 66^\circ\sin 6^\circ)(2\sin 78^\circ\sin 42^\circ)}{(2\cos 66^\circ\cos 6^\circ)(2\cos 78^\circ\cos 42^\circ)} \)
\( = \frac{(\cos 60^\circ - \cos 72^\circ)(\cos 36^\circ - \cos 120^\circ)}{(\cos 60^\circ + \cos 72^\circ)(\cos 36^\circ + \cos 120^\circ)} = \frac{(\cos 60^\circ - \sin 18^\circ)(\cos 36^\circ + \sin 30^\circ)}{(\cos 60^\circ + \sin 18^\circ)(\cos 36^\circ - \sin 30^\circ)} \)
\( = \frac{\left(\frac{1}{2} - \frac{\sqrt{5} - 1}{4}\right)\left(\frac{\sqrt{5} + 1}{4} + \frac{1}{2}\right)}{\left(\frac{1}{2} + \frac{\sqrt{5} - 1}{4}\right)\left(\frac{\sqrt{5} + 1}{4} - \frac{1}{2}\right)} = \frac{(3 - \sqrt{5})(3 + \sqrt{5})}{(\sqrt{5} + 1)(\sqrt{5} - 1)} = \frac{9 - 5}{5 - 1} \) = 1 = RHS

Example 8. Prove that: \( 4\sin 27^\circ = \sqrt{5 + \sqrt{5}} - \sqrt{3 - \sqrt{5}} \)
Answer:
We have 16 sin²27° = 8 (1 − cos 54°) = 8 (1 − sin 36°)
\( \Rightarrow 16\sin^2 27^\circ = 8\left\{1 - \frac{\sqrt{10 - 2\sqrt{5}}}{4}\right\} = 2\left\{4 - \sqrt{10 - 2\sqrt{5}}\right\} = 8 - 2\sqrt{10 - 2\sqrt{5}} \)
\( \Rightarrow 16\sin^2 27^\circ = (5 + \sqrt{5}) + (3 - \sqrt{5}) - 2\sqrt{(5 + \sqrt{5})(3 - \sqrt{5})} \)
\( \Rightarrow 16\sin^2 27^\circ = \left\{\sqrt{5 + \sqrt{5}}\right\}^2 + \left\{\sqrt{3 - \sqrt{5}}\right\}^2 - 2\sqrt{(5 + \sqrt{5})(3 - \sqrt{5})} = \left\{\sqrt{5 + \sqrt{5}} - \sqrt{3 - \sqrt{5}}\right\}^2 \)
Since sin 27° > 0 and \( \sqrt{5 + \sqrt{5}} > \sqrt{3 - \sqrt{5}} \), taking the positive square root: \( 4\sin 27^\circ = \sqrt{5 + \sqrt{5}} - \sqrt{3 - \sqrt{5}} \)

Example 9. Prove that \( \frac{\tan 3x}{\tan x} \) never lies between \( \frac{1}{3} \) and 3.
Answer:
Let \( y = \frac{\tan 3x}{\tan x} \). Then \( y = \frac{3\tan x - \tan^3 x}{\tan x(1 - 3\tan^2 x)} = \frac{3 - \tan^2 x}{1 - 3\tan^2 x} \)
⇒ (3y − 1) tan²x = y − 3 ⇒ \( \tan^2 x = \frac{y - 3}{3y - 1} \)
Now, tan²x ≥ 0 for all x, ∴ \( \frac{y - 3}{3y - 1} \ge 0 \) ⇒ \( \frac{(y - 3)(3y - 1)}{(3y - 1)^2} \ge 0 \)  [∵ (3y − 1)² > 0]
⇒ (y − 3)(3y − 1) ≥ 0 ⇒ \( y \le \frac{1}{3} \) or y ≥ 3
⇒ y does not lie between 1/3 and 3.

Example 10. If A + B + C = π, then prove the following:
(i) sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C
(ii) sin²A + sin²B + sin²C = 2 + 2 cos A cos B cos C
(iii) \( \cos^2\frac{A}{2} + \cos^2\frac{B}{2} + \cos^2\frac{C}{2} = 2 + 2\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2} \)
(iv) \( \cos\frac{A}{2} + \cos\frac{B}{2} + \cos\frac{C}{2} = 4\cos\frac{\pi - A}{4}\cos\frac{\pi - B}{4}\cos\frac{\pi - C}{4} \)
(v) \( \frac{\cos A - \cos B + \cos C + 1}{\cos A + \cos B + \cos C - 1} = \cot\frac{A}{2}\cot\frac{C}{2} \)
Answer:
(i) L.H.S. = sin 2A + sin 2B + sin 2C = 2 sin (A + B) cos (A − B) + 2 sin C cos C
= 2 sin C cos (A − B) − 2 sin C cos (A + B)  [∵ sin (A + B) = sin C, cos C = −cos (A + B)]
= 2 sin C [cos (A − B) − cos (A + B)] = 2 sin C × 2 sin A sin B = 4 sin A sin B sin C = R.H.S.
(ii) L.H.S. \( = \sin^2 A + \frac{1 - \cos 2B}{2} + \frac{1 - \cos 2C}{2} = \frac{1}{2}[2 + 2\sin^2 A - (\cos 2B + \cos 2C)] \)
= 1 + sin²A − cos (B + C) cos (B − C) = 2 − cos²A + cos A cos (B − C)  [∵ cos (B + C) = −cos A]
= 2 + cos A [−cos A + cos (B − C)] = 2 + cos A [cos (B + C) + cos (B − C)] = 2 + cos A × 2 cos B cos C = 2 + 2 cos A cos B cos C = R.H.S.
(iii) L.H.S. \( = \cos^2\frac{A}{2} + \frac{1 + \cos B}{2} + \frac{1 + \cos C}{2} = 1 + \cos^2\frac{A}{2} + \frac{1}{2}(\cos B + \cos C) \)
\( = 1 + \cos^2\frac{A}{2} + \cos\frac{B + C}{2}\cos\frac{B - C}{2} = 2 - \sin^2\frac{A}{2} + \sin\frac{A}{2}\cos\frac{B - C}{2} \)  [∵ \( \cos\frac{B + C}{2} = \sin\frac{A}{2} \)]
\( = 2 - \sin\frac{A}{2}\left[\cos\frac{B + C}{2} - \cos\frac{B - C}{2}\right] = 2 - \sin\frac{A}{2}\left[-2\sin\frac{B}{2}\sin\frac{C}{2}\right] = 2 + 2\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2} \) = R.H.S.
(iv) L.H.S. \( = \sin\frac{\pi - A}{2} + 2\cos\frac{B + C}{4}\cos\frac{B - C}{4} = 2\sin\frac{\pi - A}{4}\cos\frac{\pi - A}{4} + 2\cos\frac{\pi - A}{4}\cos\frac{B - C}{4} \)  [∵ \( \frac{B + C}{4} = \frac{\pi - A}{4} \)]
\( = 2\cos\frac{\pi - A}{4}\left[\sin\frac{B + C}{4} + \cos\frac{B - C}{4}\right] = 2\cos\frac{\pi - A}{4}\left[\cos\left(\frac{\pi - B}{4} + \frac{\pi - C}{4}\right) + \cos\left(\frac{\pi - C}{4} - \frac{\pi - B}{4}\right)\right] \)
\( = 2\cos\frac{\pi - A}{4} \cdot 2\cos\frac{\pi - B}{4}\cos\frac{\pi - C}{4} = 4\cos\frac{\pi - A}{4}\cos\frac{\pi - B}{4}\cos\frac{\pi - C}{4} \) = R.H.S.
(v) L.H.S. \( = \frac{(\cos A + \cos C) + (1 - \cos B)}{(\cos A + \cos C) - (1 - \cos B)} = \frac{2\cos\frac{A + C}{2}\cos\frac{A - C}{2} + 2\sin^2\frac{B}{2}}{2\cos\frac{A + C}{2}\cos\frac{A - C}{2} - 2\sin^2\frac{B}{2}} \)
\( = \frac{2\cos\frac{A + C}{2}\left[\cos\frac{A - C}{2} + \cos\frac{A + C}{2}\right]}{2\cos\frac{A + C}{2}\left[\cos\frac{A - C}{2} - \cos\frac{A + C}{2}\right]} \)  [∵ \( \sin\frac{B}{2} = \cos\frac{A + C}{2} \)]
\( = \frac{2\cos\frac{A}{2}\cos\frac{C}{2}}{2\sin\frac{A}{2}\sin\frac{C}{2}} = \cot\frac{A}{2}\cot\frac{C}{2} \) = R.H.S.