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Punjab Board Class 11 Mathematics

Exercise 10.1 Solutions — Trigonometric Identities

Complete worked solutions for Exercise 10.1 from Chapter 10 (Trigonometric Identities), Punjab Board Class 11 Mathematics. Each question shows every step and names the formula used.

Mathematics 11 (Punjab Board)

Exercise 10.1 Solutions

Worked solutions for Punjab Board Class 11 Mathematics, Chapter 10: Trigonometric Identities.

Question 1: Find the values of: (i) \(\cos(-1230^\circ)\), (ii) \(\tan(-1035^\circ)\), (iii) \(\sec(1140^\circ)\), (iv) \(\csc(-690^\circ)\), (v) \(\cot(1320^\circ)\), (vi) \(\cos(-240^\circ)\)

Bade angles ko pehle \(360^\circ\) ke multiples subtract kar ke chota karenge, phir allied angles ki identities apply karenge. Negative angles ke liye pehle sign hata lenge (cos aur sec even hain, baqi odd).

(i) \(\cos(-1230^\circ) = \cos 1230^\circ\) (cosine is even). Since \(1230^\circ = 3 \times 360^\circ + 150^\circ\):

\[ \cos 1230^\circ = \cos 150^\circ = \cos(180^\circ - 30^\circ) = -\cos 30^\circ = -\frac{\sqrt{3}}{2} \]

(ii) \(\tan(-1035^\circ) = -\tan 1035^\circ\). Since \(1035^\circ = 2 \times 360^\circ + 315^\circ\):

\[ -\tan 1035^\circ = -\tan 315^\circ = -\tan(360^\circ - 45^\circ) = -(-\tan 45^\circ) = 1 \]

(iii) \(1140^\circ = 3 \times 360^\circ + 60^\circ\):

\[ \sec 1140^\circ = \frac{1}{\cos 1140^\circ} = \frac{1}{\cos 60^\circ} = \frac{1}{1/2} = 2 \]

(iv) \(\csc(-690^\circ) = -\csc 690^\circ\). Since \(690^\circ = 360^\circ + 330^\circ\):

\[ -\csc 690^\circ = -\csc 330^\circ = -\csc(360^\circ - 30^\circ) = -(-\csc 30^\circ) = 2 \]

(v) \(1320^\circ = 3 \times 360^\circ + 240^\circ\):

\[ \cot 1320^\circ = \cot 240^\circ = \cot(180^\circ + 60^\circ) = \cot 60^\circ = \frac{1}{\sqrt{3}} \]

(vi) \(\cos(-240^\circ) = \cos 240^\circ\):

\[ \cos 240^\circ = \cos(180^\circ + 60^\circ) = -\cos 60^\circ = -\frac{1}{2} \]
Answer: (i) \(-\frac{\sqrt{3}}{2}\), (ii) \(1\), (iii) \(2\), (iv) \(2\), (v) \(\frac{1}{\sqrt{3}}\), (vi) \(-\frac{1}{2}\).

Question 2: Express in terms of a trigonometric ratio of an acute angle: (i) \(\cos 168^\circ\), (ii) \(\sin 192^\circ\), (iii) \(\cos 333^\circ\), (iv) \(\tan 213^\circ\), (v) \(\cos(-435^\circ)\), (vi) \(\sin 219^\circ\), (vii) \(\tan(-597^\circ)\), (viii) \(\csc(-111^\circ)\), (ix) \(\sin(-390^\circ)\)

Har angle ka quadrant maloom kar ke allied angle identity lagayenge. Acute angle ka matlab \(0^\circ\) se \(90^\circ\) ke darmiyan ka angle hai.

(i) \(\cos 168^\circ = \cos(180^\circ - 12^\circ) = -\cos 12^\circ\)

(ii) \(\sin 192^\circ = \sin(180^\circ + 12^\circ) = -\sin 12^\circ\)

(iii) \(\cos 333^\circ = \cos(360^\circ - 27^\circ) = \cos 27^\circ\)

(iv) \(\tan 213^\circ = \tan(180^\circ + 33^\circ) = \tan 33^\circ\)

(v) \(\cos(-435^\circ) = \cos 435^\circ = \cos(360^\circ + 75^\circ) = \cos 75^\circ\)

(vi) \(\sin 219^\circ = \sin(180^\circ + 39^\circ) = -\sin 39^\circ\)

(vii) \(\tan(-597^\circ) = -\tan 597^\circ = -\tan(360^\circ + 237^\circ) = -\tan 237^\circ = -\tan(180^\circ + 57^\circ) = -\tan 57^\circ\)

(viii) \(\csc(-111^\circ) = -\csc 111^\circ = -\csc(180^\circ - 69^\circ) = -\csc 69^\circ\)

(ix) \(\sin(-390^\circ) = -\sin 390^\circ = -\sin(360^\circ + 30^\circ) = -\sin 30^\circ = -\frac{1}{2}\)

Common mistake: Negative angle mein sign ghalat lagana. Pehle angle ko positive banao (odd/even function ke hisab se), phir quadrant se sign decide karo.

Answer: (i) \(-\cos 12^\circ\), (ii) \(-\sin 12^\circ\), (iii) \(\cos 27^\circ\), (iv) \(\tan 33^\circ\), (v) \(\cos 75^\circ\), (vi) \(-\sin 39^\circ\), (vii) \(-\tan 57^\circ\), (viii) \(-\csc 69^\circ\), (ix) \(-\frac{1}{2}\).

Question 3: Prove that

L.H.S se start kar ke allied angles ki identities apply karenge aur R.H.S tak pohnchenge.

(i) \(\sin(180^\circ + \alpha)\sin(90^\circ - \alpha) = -\sin\alpha\cos\alpha\)

\[ \text{L.H.S} = (-\sin\alpha)(\cos\alpha) = -\sin\alpha\cos\alpha = \text{R.H.S} \]

(ii) \(\sin 810^\circ\cos 630^\circ + \cos 135^\circ\sin 225^\circ = -\frac{1}{2}\)

\(\sin 810^\circ = \sin(2 \times 360^\circ + 90^\circ) = 1\); \(\cos 630^\circ = \cos(360^\circ + 270^\circ) = \cos 270^\circ = 0\); \(\cos 135^\circ = -\frac{1}{\sqrt{2}}\); \(\sin 225^\circ = -\frac{1}{\sqrt{2}}\).

\[ \text{L.H.S} = (1)(0) + \left(-\frac{1}{\sqrt{2}}\right)\left(-\frac{1}{\sqrt{2}}\right) = \frac{1}{2} \]

(iii) \(\tan 150^\circ\cot 330^\circ - 2\sec 135^\circ\csc 225^\circ = -3\)

\(\tan 150^\circ = -\tan 30^\circ = -\frac{1}{\sqrt{3}}\); \(\cot 330^\circ = -\cot 30^\circ = -\sqrt{3}\); \(\sec 135^\circ = -\sqrt{2}\); \(\csc 225^\circ = -\sqrt{2}\).

\[ \text{L.H.S} = \left(-\frac{1}{\sqrt{3}}\right)(-\sqrt{3}) - 2(-\sqrt{2})(-\sqrt{2}) = 1 - 4 = -3 = \text{R.H.S} \]

(iv) \(\sin 210^\circ + \cos 240^\circ + \tan 225^\circ + \cot 225^\circ = 1\)

\[ \text{L.H.S} = \left(-\frac{1}{2}\right) + \left(-\frac{1}{2}\right) + 1 + 1 = 1 = \text{R.H.S} \]

Common mistake: Part (ii) ka durust jawab \(+\frac{1}{2}\) hai — printed key mein likha \(-\frac{1}{2}\) ghalat hai.

Answer: (i) proved; (ii) \(= \frac{1}{2}\) (not \(-\frac{1}{2}\)); (iii) proved; (iv) proved.

Question 4: Prove that (i) \(\frac{\tan(180^\circ + \alpha)\cot(90^\circ - \alpha)}{\sin(360^\circ - \alpha)\cos(270^\circ + \alpha)} = -\sec^2\alpha\), (ii) \(\frac{\sin^2(\pi + \theta)\tan\left(\frac{3\pi}{2} + \theta\right)}{\cot^2\left(\frac{3\pi}{2} - \theta\right)\cos^2(\pi - \theta)\csc(2\pi - \theta)} = \cos\theta\), (iii) \(\frac{\cos(90^\circ + \theta)\sec(-\theta)\tan(180^\circ - \theta)}{\sec(360^\circ - \theta)\sin(180^\circ + \theta)\cot(90^\circ - \theta)} = -1\)

Numerator aur denominator ke har factor ko allied angle identities se simplify karenge, phir cancel karenge.

(i)

\[ \text{L.H.S} = \frac{(\tan\alpha)(\tan\alpha)}{(-\sin\alpha)(\sin\alpha)} = \frac{\tan^2\alpha}{-\sin^2\alpha} = -\frac{1}{\cos^2\alpha} = -\sec^2\alpha = \text{R.H.S} \]

(ii)

Numerator: \(\sin^2(\pi+\theta) = \sin^2\theta\); \(\tan\left(\frac{3\pi}{2}+\theta\right) = -\cot\theta\). Denominator: \(\cot^2\left(\frac{3\pi}{2}-\theta\right) = \tan^2\theta\); \(\cos^2(\pi-\theta) = \cos^2\theta\); \(\csc(2\pi-\theta) = -\csc\theta\).

\[ \text{L.H.S} = \frac{\sin^2\theta \cdot (-\cot\theta)}{\tan^2\theta \cdot \cos^2\theta \cdot (-\csc\theta)} = \frac{-\sin\theta\cos\theta}{-\tan^2\theta\cos^2\theta\csc\theta} = \cos\theta = \text{R.H.S} \]

(iii)

Numerator: \(\cos(90^\circ+\theta) = -\sin\theta\); \(\sec(-\theta) = \sec\theta\); \(\tan(180^\circ-\theta) = -\tan\theta\). Denominator: \(\sec(360^\circ-\theta) = \sec\theta\); \(\sin(180^\circ+\theta) = -\sin\theta\); \(\cot(90^\circ-\theta) = \tan\theta\).

\[ \text{L.H.S} = \frac{(-\sin\theta)(\sec\theta)(-\tan\theta)}{(\sec\theta)(-\sin\theta)(\tan\theta)} = -1 = \text{R.H.S} \]
Answer: All three identities are proved.

Question 5: Show that \(\sec\left(\frac{3\pi}{2} - \theta\right)\sec\left(\frac{5\pi}{2} - \theta\right) - \tan\left(\frac{3\pi}{2} - \theta\right)\tan\left(\frac{5\pi}{2} + \theta\right) = -1\)

\(\frac{5\pi}{2} = 2\pi + \frac{\pi}{2}\) hone ki wajah se us mein \(2\pi\) subtract kar ke chota angle banao, phir allied angle identities lagao.

Simplify each factor: \(\sec\left(\frac{3\pi}{2}-\theta\right) = -\csc\theta\); \(\sec\left(\frac{5\pi}{2}-\theta\right) = \sec\left(\frac{\pi}{2}-\theta\right) = \csc\theta\); \(\tan\left(\frac{3\pi}{2}-\theta\right) = \cot\theta\); \(\tan\left(\frac{5\pi}{2}+\theta\right) = \tan\left(\frac{\pi}{2}+\theta\right) = -\cot\theta\).

\[ \text{L.H.S} = (-\csc\theta)(\csc\theta) - (\cot\theta)(-\cot\theta) = -\csc^2\theta + \cot^2\theta \]
\[ = -(\csc^2\theta - \cot^2\theta) = -1 = \text{R.H.S} \]
Answer: The identity is proved using \(\csc^2\theta - \cot^2\theta = 1\).

Question 6: If \(\alpha + \beta + \gamma = 180^\circ\), show that (i) \(\sin(\alpha+\beta) = \sin\gamma\), (ii) \(\sec\left(\frac{\alpha+\beta}{2}\right) = \csc\left(\frac{\gamma}{2}\right)\), (iii) \(\csc\alpha = \frac{1}{\sin(\beta+\gamma)}\), (iv) \(\tan(\alpha+\beta) + \tan\gamma = 0\)

\(\alpha + \beta = 180^\circ - \gamma\) likh kar allied angle identities apply karenge.

(i)

\[ \sin(\alpha+\beta) = \sin(180^\circ - \gamma) = \sin\gamma \]

(ii)

\[ \sec\left(\frac{\alpha+\beta}{2}\right) = \sec\left(\frac{180^\circ-\gamma}{2}\right) = \sec\left(90^\circ - \frac{\gamma}{2}\right) = \csc\left(\frac{\gamma}{2}\right) \]

(iii)

\[ \frac{1}{\sin(\beta+\gamma)} = \frac{1}{\sin(180^\circ-\alpha)} = \frac{1}{\sin\alpha} = \csc\alpha \]

(iv)

\[ \tan(\alpha+\beta) + \tan\gamma = \tan(180^\circ-\gamma) + \tan\gamma = -\tan\gamma + \tan\gamma = 0 \]
Answer: All four relations are proved.