Mathematics 11 (Punjab Board)
Exercise 10.2 Solutions
Worked solutions for Punjab Board Class 11 Mathematics, Chapter 10: Trigonometric Identities.
Question 1: Without using table find the values of the following: (i) \(\sin 15^\circ\), (ii) \(\cos 15^\circ\), (iii) \(\tan 15^\circ\), (iv) \(\sin 105^\circ\), (v) \(\cos 105^\circ\), (vi) \(\tan 105^\circ\)
Angles ko special angles ke sum ya difference mein likh kar compound angle formulas lagayenge. \(\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}\), \(\sin 30^\circ = \frac{1}{2}\), \(\cos 30^\circ = \frac{\sqrt{3}}{2}\) yaad rakhen.
(i) \(\sin 15^\circ = \sin(45^\circ - 30^\circ) = \sin 45^\circ\cos 30^\circ - \cos 45^\circ\sin 30^\circ\)
\[
= \frac{1}{\sqrt{2}}\cdot\frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}}\cdot\frac{1}{2} = \frac{\sqrt{3}-1}{2\sqrt{2}}
\]
(ii) \(\cos 15^\circ = \cos(45^\circ - 30^\circ) = \cos 45^\circ\cos 30^\circ + \sin 45^\circ\sin 30^\circ\)
\[
= \frac{1}{\sqrt{2}}\cdot\frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}}\cdot\frac{1}{2} = \frac{\sqrt{3}+1}{2\sqrt{2}}
\]
(iii) \(\tan 15^\circ = \tan(45^\circ - 30^\circ) = \frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}}\)
\[
= \frac{\sqrt{3}-1}{\sqrt{3}+1} = \frac{(\sqrt{3}-1)^2}{3-1} = \frac{4-2\sqrt{3}}{2} = 2-\sqrt{3}
\]
(iv) \(\sin 105^\circ = \sin(60^\circ + 45^\circ) = \sin 60^\circ\cos 45^\circ + \cos 60^\circ\sin 45^\circ\)
\[
= \frac{\sqrt{3}}{2}\cdot\frac{1}{\sqrt{2}} + \frac{1}{2}\cdot\frac{1}{\sqrt{2}} = \frac{\sqrt{3}+1}{2\sqrt{2}}
\]
(v) \(\cos 105^\circ = \cos(60^\circ + 45^\circ) = \cos 60^\circ\cos 45^\circ - \sin 60^\circ\sin 45^\circ\)
\[
= \frac{1}{2}\cdot\frac{1}{\sqrt{2}} - \frac{\sqrt{3}}{2}\cdot\frac{1}{\sqrt{2}} = \frac{1-\sqrt{3}}{2\sqrt{2}}
\]
(vi) \(\tan 105^\circ = \tan(180^\circ - 75^\circ) = -\tan 75^\circ\)
\[
\tan 75^\circ = \frac{1 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}} = \frac{\sqrt{3}+1}{\sqrt{3}-1} = 2+\sqrt{3}, \quad \text{so } \tan 105^\circ = -(2+\sqrt{3})
\]
Answer: (i) \(\frac{\sqrt{3}-1}{2\sqrt{2}}\), (ii) \(\frac{\sqrt{3}+1}{2\sqrt{2}}\), (iii) \(2-\sqrt{3}\), (iv) \(\frac{\sqrt{3}+1}{2\sqrt{2}}\), (v) \(\frac{1-\sqrt{3}}{2\sqrt{2}}\), (vi) \(-(2+\sqrt{3})\).
Question 2: Prove that: (i) \(\sin(45^\circ + \alpha) = \frac{1}{\sqrt{2}}(\sin\alpha + \cos\alpha)\), (ii) \(\cos(\alpha + 45^\circ) = \frac{1}{\sqrt{2}}(\cos\alpha - \sin\alpha)\)
L.H.S par addition formulas lagayenge aur \(\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}\) substitute karenge.
(i)
\[
\text{L.H.S} = \sin 45^\circ\cos\alpha + \cos 45^\circ\sin\alpha = \frac{1}{\sqrt{2}}\cos\alpha + \frac{1}{\sqrt{2}}\sin\alpha = \frac{1}{\sqrt{2}}(\sin\alpha + \cos\alpha) = \text{R.H.S}
\]
(ii)
\[
\text{L.H.S} = \cos\alpha\cos 45^\circ - \sin\alpha\sin 45^\circ = \frac{1}{\sqrt{2}}\cos\alpha - \frac{1}{\sqrt{2}}\sin\alpha = \frac{1}{\sqrt{2}}(\cos\alpha - \sin\alpha) = \text{R.H.S}
\]
Answer: Both identities are proved.
Question 3: Prove that: (i) \(\tan(45^\circ + A)\tan(45^\circ - A) = 1\), (ii) \(\tan\left(\frac{\pi}{4} - \theta\right) + \tan\left(\frac{3\pi}{4} + \theta\right) = 0\), (iii) \(\sin\left(\theta + \frac{\pi}{6}\right) + \cos\left(\theta + \frac{\pi}{3}\right) = \cos\theta\), (iv) \(\frac{\sin\theta - \cos\theta\tan\frac{\theta}{2}}{\cos\theta + \sin\theta\tan\frac{\theta}{2}} = \tan\frac{\theta}{2}\), (v) \(\frac{1 - \tan\theta\tan\phi}{1 + \tan\theta\tan\phi} = \frac{\cos(\theta+\phi)}{\cos(\theta-\phi)}\)
Har part mein pehle tan ya compound angle ke formulas lagayenge, phir simplify kar ke R.H.S tak pohnchenge.
(i)
\[
\text{L.H.S} = \left(\frac{1+\tan A}{1-\tan A}\right)\left(\frac{1-\tan A}{1+\tan A}\right) = 1 = \text{R.H.S}
\]
(ii) Since \(\tan\left(\frac{3\pi}{4}+\theta\right) = \tan\left(\pi - \left(\frac{\pi}{4}-\theta\right)\right) = -\tan\left(\frac{\pi}{4}-\theta\right)\):
\[
\text{L.H.S} = \tan\left(\frac{\pi}{4}-\theta\right) - \tan\left(\frac{\pi}{4}-\theta\right) = 0 = \text{R.H.S}
\]
(iii)
\[
\text{L.H.S} = \sin\theta\cos\frac{\pi}{6} + \cos\theta\sin\frac{\pi}{6} + \cos\theta\cos\frac{\pi}{3} - \sin\theta\sin\frac{\pi}{3}
\]
\[
= \sin\theta\left(\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2}\right) + \cos\theta\left(\frac{1}{2} + \frac{1}{2}\right) = \cos\theta = \text{R.H.S}
\]
(iv) Dividing numerator and denominator by \(\cos\theta\):
\[
\text{L.H.S} = \frac{\tan\theta - \tan\frac{\theta}{2}}{1 + \tan\theta\tan\frac{\theta}{2}}
\]
Put \(t = \tan\frac{\theta}{2}\) and use \(\tan\theta = \frac{2t}{1-t^2}\):
\[
= \frac{\frac{2t}{1-t^2} - t}{1 + \frac{2t^2}{1-t^2}} = \frac{t(1+t^2)}{1+t^2} = t = \tan\frac{\theta}{2} = \text{R.H.S}
\]
(v) Dividing numerator and denominator by \(\cos\theta\cos\phi\):
\[
\text{L.H.S} = \frac{\frac{\cos\theta\cos\phi - \sin\theta\sin\phi}{\cos\theta\cos\phi}}{\frac{\cos\theta\cos\phi + \sin\theta\sin\phi}{\cos\theta\cos\phi}} = \frac{\cos(\theta+\phi)}{\cos(\theta-\phi)} = \text{R.H.S}
\]
Answer: All five identities are proved.
Question 4: Show that \(\cos(\alpha+\beta)\cos(\alpha-\beta) = \cos^2\alpha - \sin^2\beta = \cos^2\beta - \sin^2\alpha\)
Dono factors ko expand kar ke difference of squares banayenge, phir \(\cos^2 = 1 - \sin^2\) ki substitution se simplify karenge.
\[
\text{L.H.S} = (\cos\alpha\cos\beta - \sin\alpha\sin\beta)(\cos\alpha\cos\beta + \sin\alpha\sin\beta) = \cos^2\alpha\cos^2\beta - \sin^2\alpha\sin^2\beta
\]
\[
= \cos^2\alpha(1-\sin^2\beta) - (1-\cos^2\alpha)\sin^2\beta = \cos^2\alpha - \sin^2\beta
\]
By symmetry (interchanging \(\alpha\) and \(\beta\) in the same steps), it also equals \(\cos^2\beta - \sin^2\alpha\).
Answer: \(\cos(\alpha+\beta)\cos(\alpha-\beta) = \cos^2\alpha - \sin^2\beta = \cos^2\beta - \sin^2\alpha\). Proved.
Question 5: Show that \(\frac{\sin(\alpha+\beta) + \sin(\alpha-\beta)}{\cos(\alpha+\beta) + \cos(\alpha-\beta)} = \tan\alpha\)
Numerator aur denominator dono ko expand karenge — sinβ aur cosβ wale terms cancel ho jayenge.
\[
\text{L.H.S} = \frac{(\sin\alpha\cos\beta + \cos\alpha\sin\beta) + (\sin\alpha\cos\beta - \cos\alpha\sin\beta)}{(\cos\alpha\cos\beta - \sin\alpha\sin\beta) + (\cos\alpha\cos\beta + \sin\alpha\sin\beta)}
\]
\[
= \frac{2\sin\alpha\cos\beta}{2\cos\alpha\cos\beta} = \frac{\sin\alpha}{\cos\alpha} = \tan\alpha = \text{R.H.S}
\]
Answer: The identity is proved.
Question 6: Show that: (i) \(\sin^2\left(\frac{\alpha+\beta}{2}\right) - \sin^2\left(\frac{\alpha-\beta}{2}\right) = \sin\alpha\sin\beta\), (ii) \(\sin^2\alpha + \sin^2\beta + \cos^2(\alpha+\beta) + 2\sin\alpha\sin\beta\cos(\alpha+\beta) = 1\)
Part (i) mein difference of squares ko factor kar ke sum-to-product formulas lagayenge. Part (ii) mein \(\cos(\alpha+\beta)\) ko expand kar ke simplify karenge.
(i)
\[
\text{L.H.S} = \left[\sin\frac{\alpha+\beta}{2} - \sin\frac{\alpha-\beta}{2}\right]\left[\sin\frac{\alpha+\beta}{2} + \sin\frac{\alpha-\beta}{2}\right]
\]
\[
= \left[2\cos\frac{\alpha}{2}\sin\frac{\beta}{2}\right]\left[2\sin\frac{\alpha}{2}\cos\frac{\beta}{2}\right] = \sin\alpha\sin\beta = \text{R.H.S}
\]
(ii) Expand \(\cos^2(\alpha+\beta) = (\cos\alpha\cos\beta - \sin\alpha\sin\beta)^2\):
\[
\text{L.H.S} = \sin^2\alpha + \sin^2\beta + \cos^2\alpha\cos^2\beta - 2\sin\alpha\sin\beta\cos\alpha\cos\beta + \sin^2\alpha\sin^2\beta + 2\sin\alpha\sin\beta(\cos\alpha\cos\beta - \sin\alpha\sin\beta)
\]
\[
= \sin^2\alpha + \sin^2\beta + \cos^2\alpha\cos^2\beta - \sin^2\alpha\sin^2\beta
\]
\[
= \sin^2\alpha(1-\sin^2\beta) + \sin^2\beta + \cos^2\alpha\cos^2\beta = \cos^2\beta(\sin^2\alpha + \cos^2\alpha) + \sin^2\beta = 1 = \text{R.H.S}
\]
Answer: Both identities are proved.
Question 7: Show that: (i) \(\cos(\alpha-\beta) = \frac{1+\tan\alpha\tan\beta}{\sec\alpha\sec\beta}\), (ii) \(\sin(\alpha+\beta) = \frac{1+\cot\alpha\tan\beta}{\csc\alpha\sec\beta}\), (iii) \(\cot(\alpha-\beta) = \frac{\cot\alpha\cot\beta+1}{\cot\beta-\cot\alpha}\), (iv) \(\frac{\tan\alpha+\tan\beta}{\tan\alpha-\tan\beta} = \frac{\sin(\alpha+\beta)}{\sin(\alpha-\beta)}\), (v) \(\cot(\alpha+\beta) = \frac{\cot\alpha\cot\beta-1}{\cot\alpha+\cot\beta}\)
R.H.S ko simplify kar ke L.H.S tak pohnchenge — tan, cot, sec, csc ko sin aur cos mein likhna aam trick hai.
(i)
\[
\text{R.H.S} = \left(1 + \frac{\sin\alpha\sin\beta}{\cos\alpha\cos\beta}\right)\cos\alpha\cos\beta = \cos\alpha\cos\beta + \sin\alpha\sin\beta = \cos(\alpha-\beta) = \text{L.H.S}
\]
(ii)
\[
\text{R.H.S} = \left(1 + \frac{\cos\alpha}{\sin\alpha}\cdot\frac{\sin\beta}{\cos\beta}\right)\sin\alpha\cos\beta = \sin\alpha\cos\beta + \cos\alpha\sin\beta = \sin(\alpha+\beta) = \text{L.H.S}
\]
(iii) Write \(\cot(\alpha-\beta) = \frac{1}{\tan(\alpha-\beta)}\):
\[
= \frac{1 + \tan\alpha\tan\beta}{\tan\alpha - \tan\beta} = \frac{1 + \frac{1}{\cot\alpha\cot\beta}}{\frac{1}{\cot\alpha} - \frac{1}{\cot\beta}} = \frac{\cot\alpha\cot\beta + 1}{\cot\beta - \cot\alpha} = \text{R.H.S}
\]
(iv)
\[
\text{L.H.S} = \frac{\frac{\sin\alpha}{\cos\alpha} + \frac{\sin\beta}{\cos\beta}}{\frac{\sin\alpha}{\cos\alpha} - \frac{\sin\beta}{\cos\beta}} = \frac{\sin\alpha\cos\beta + \cos\alpha\sin\beta}{\sin\alpha\cos\beta - \cos\alpha\sin\beta} = \frac{\sin(\alpha+\beta)}{\sin(\alpha-\beta)} = \text{R.H.S}
\]
(v) Write \(\cot(\alpha+\beta) = \frac{1}{\tan(\alpha+\beta)}\):
\[
= \frac{1 - \tan\alpha\tan\beta}{\tan\alpha + \tan\beta} = \frac{1 - \frac{1}{\cot\alpha\cot\beta}}{\frac{1}{\cot\alpha} + \frac{1}{\cot\beta}} = \frac{\cot\alpha\cot\beta - 1}{\cot\alpha + \cot\beta} = \text{R.H.S}
\]
Answer: All five identities are proved.
Question 8: If \(\sin\alpha = \frac{24}{25}\) and \(\cos\beta = \frac{20}{29}\), where \(0 < \alpha < \frac{\pi}{2}\) and \(0 < \beta < \frac{\pi}{2}\), show that \(\sin(\alpha-\beta) = \frac{333}{725}\)
Dono angles pehle quadrant mein hain, is liye \(\cos\alpha\) aur \(\sin\beta\) positive honge. Pythagorean identity \(\sin^2 + \cos^2 = 1\) se inhein nikalenge, phir \(\sin(\alpha-\beta)\) ka formula lagayenge.
\[
\cos\alpha = \sqrt{1 - \left(\frac{24}{25}\right)^2} = \sqrt{\frac{49}{625}} = \frac{7}{25}, \qquad \sin\beta = \sqrt{1 - \left(\frac{20}{29}\right)^2} = \sqrt{\frac{441}{841}} = \frac{21}{29}
\]
\[
\sin(\alpha-\beta) = \sin\alpha\cos\beta - \cos\alpha\sin\beta = \frac{24}{25}\cdot\frac{20}{29} - \frac{7}{25}\cdot\frac{21}{29} = \frac{480-147}{725} = \frac{333}{725}
\]
Answer: \(\sin(\alpha-\beta) = \frac{333}{725}\). Proved.
Question 9: If \(\sin\alpha = -\frac{8}{17}\) and \(\cos\beta = -\frac{4}{5}\), where \(\frac{3\pi}{2} < \alpha < 2\pi\) and \(\pi < \beta < \frac{3\pi}{2}\), find (i) \(\sin(\alpha+\beta)\), (ii) \(\cos(\alpha+\beta)\), (iii) \(\tan(\alpha+\beta)\), (iv) \(\sin(\alpha-\beta)\), (v) \(\cos(\alpha-\beta)\), (vi) \(\tan(\alpha-\beta)\). Also determine the quadrant of \(\alpha+\beta\) and \(\alpha-\beta\).
\(\alpha\) fourth quadrant mein hai (cos positive) aur \(\beta\) third quadrant mein hai (sin negative). Pythagorean identity se \(\cos\alpha = \frac{15}{17}\) aur \(\sin\beta = -\frac{3}{5}\) niklega.
\[
\cos\alpha = \sqrt{1 - \frac{64}{289}} = \frac{15}{17}, \qquad \sin\beta = -\sqrt{1 - \frac{16}{25}} = -\frac{3}{5}
\]
(i) \(\sin(\alpha+\beta) = \left(-\frac{8}{17}\right)\left(-\frac{4}{5}\right) + \left(\frac{15}{17}\right)\left(-\frac{3}{5}\right) = \frac{32-45}{85} = -\frac{13}{85}\)
(ii) \(\cos(\alpha+\beta) = \left(\frac{15}{17}\right)\left(-\frac{4}{5}\right) - \left(-\frac{8}{17}\right)\left(-\frac{3}{5}\right) = \frac{-60-24}{85} = -\frac{84}{85}\)
(iii) \(\tan(\alpha+\beta) = \frac{-13/85}{-84/85} = \frac{13}{84}\)
(iv) \(\sin(\alpha-\beta) = \left(-\frac{8}{17}\right)\left(-\frac{4}{5}\right) - \left(\frac{15}{17}\right)\left(-\frac{3}{5}\right) = \frac{32+45}{85} = \frac{77}{85}\)
(v) \(\cos(\alpha-\beta) = \left(\frac{15}{17}\right)\left(-\frac{4}{5}\right) + \left(-\frac{8}{17}\right)\left(-\frac{3}{5}\right) = \frac{-60+24}{85} = -\frac{36}{85}\)
(vi) \(\tan(\alpha-\beta) = \frac{77/85}{-36/85} = -\frac{77}{36}\)
Quadrants: \(\alpha+\beta\) has \(\sin < 0\), \(\cos < 0\), \(\tan > 0\) → Quadrant III. \(\alpha-\beta\) has \(\sin > 0\), \(\cos < 0\), \(\tan < 0\) → Quadrant II.
Answer: (i) \(-\frac{13}{85}\), (ii) \(-\frac{84}{85}\), (iii) \(\frac{13}{84}\), (iv) \(\frac{77}{85}\), (v) \(-\frac{36}{85}\), (vi) \(-\frac{77}{36}\); \(\alpha+\beta\) in Quadrant III, \(\alpha-\beta\) in Quadrant II.
Question 10: Find \(\sin(\alpha+\beta)\) and \(\cos(\alpha+\beta)\), given that: (i) \(\tan\alpha = \frac{3}{4}\), \(\cos\beta = \frac{5}{13}\) and neither terminal side is in Quadrant I; (ii) \(\tan\alpha = -\frac{15}{8}\), \(\sin\beta = -\frac{7}{25}\) and neither terminal side is in Quadrant IV.
Pehle quadrant maloom karenge: tan positive aur Quadrant I nahi → Quadrant III; cos positive aur Quadrant I nahi → Quadrant IV. Signs isi hisab se lagayenge.
(i) \(\alpha\) is in Quadrant III (tan positive), \(\beta\) is in Quadrant IV (cos positive):
\[
\sin\alpha = -\frac{3}{5}, \quad \cos\alpha = -\frac{4}{5}, \quad \sin\beta = -\frac{12}{13}
\]
\[
\sin(\alpha+\beta) = \left(-\frac{3}{5}\right)\left(\frac{5}{13}\right) + \left(-\frac{4}{5}\right)\left(-\frac{12}{13}\right) = \frac{-15+48}{65} = \frac{33}{65}
\]
\[
\cos(\alpha+\beta) = \left(-\frac{4}{5}\right)\left(\frac{5}{13}\right) - \left(-\frac{3}{5}\right)\left(-\frac{12}{13}\right) = \frac{-20-36}{65} = -\frac{56}{65}
\]
Answer: (i) \(\sin(\alpha+\beta) = \frac{33}{65}\), \(\cos(\alpha+\beta) = -\frac{56}{65}\).
(ii) \(\alpha\) is in Quadrant II (tan negative, not Quadrant IV), \(\beta\) is in Quadrant III (sin negative, not Quadrant IV):
\[
\sin\alpha = \frac{15}{17}, \quad \cos\alpha = -\frac{8}{17}, \quad \cos\beta = -\frac{24}{25}
\]
\[
\sin(\alpha+\beta) = \left(\frac{15}{17}\right)\left(-\frac{24}{25}\right) + \left(-\frac{8}{17}\right)\left(-\frac{7}{25}\right) = \frac{-360+56}{425} = -\frac{304}{425}
\]
\[
\cos(\alpha+\beta) = \left(-\frac{8}{17}\right)\left(-\frac{24}{25}\right) - \left(\frac{15}{17}\right)\left(-\frac{7}{25}\right) = \frac{192+105}{425} = \frac{297}{425}
\]
Answer: (ii) \(\sin(\alpha+\beta) = -\frac{304}{425}\), \(\cos(\alpha+\beta) = \frac{297}{425}\).
Common mistake: Quadrant ghalat maloom karna. "Neither in Quadrant I" ka matlab "Quadrant I ke ilawa woh quadrant jahan given sign possible ho" hai — dono options check karo.
Question 11: Prove that \(\frac{\cos 19^\circ + \sin 19^\circ}{\cos 19^\circ - \sin 19^\circ} = \tan 64^\circ\)
Numerator aur denominator dono ko \(\cos 19^\circ\) se divide karenge taake tan ki form ban jaye, phir \(\tan(45^\circ + 19^\circ)\) ka formula pehchano.
\[
\text{L.H.S} = \frac{1 + \tan 19^\circ}{1 - \tan 19^\circ} = \frac{\tan 45^\circ + \tan 19^\circ}{1 - \tan 45^\circ\tan 19^\circ} = \tan(45^\circ + 19^\circ) = \tan 64^\circ = \text{R.H.S}
\]
Answer: The identity is proved.
Question 12: Prove that \(\cos(60^\circ+\theta)\cos(60^\circ-\theta) + \sin(60^\circ+\theta)\sin(60^\circ-\theta) = \cos 2\theta\)
Ye \(\cos(A-B) = \cos A\cos B + \sin A\sin B\) ka pattern hai — \(A = 60^\circ+\theta\), \(B = 60^\circ-\theta\) rakho.
\[
\text{L.H.S} = \cos\big[(60^\circ+\theta) - (60^\circ-\theta)\big] = \cos 2\theta = \text{R.H.S}
\]
Answer: The identity is proved.
Question 13: If \(\alpha, \beta, \gamma\) are the angles of a triangle ABC, show that \(\cot\frac{\alpha}{2} + \cot\frac{\beta}{2} + \cot\frac{\gamma}{2} = \cot\frac{\alpha}{2}\cot\frac{\beta}{2}\cot\frac{\gamma}{2}\)
\(\frac{\alpha}{2} + \frac{\beta}{2} = 90^\circ - \frac{\gamma}{2}\) se start karenge aur dono taraf cot le kar simplify karenge.
\[
\cot\left(\frac{\alpha}{2} + \frac{\beta}{2}\right) = \cot\left(90^\circ - \frac{\gamma}{2}\right) = \tan\frac{\gamma}{2} = \frac{1}{\cot\frac{\gamma}{2}}
\]
\[
\frac{\cot\frac{\alpha}{2}\cot\frac{\beta}{2} - 1}{\cot\frac{\beta}{2} + \cot\frac{\alpha}{2}} = \frac{1}{\cot\frac{\gamma}{2}}
\]
Cross-multiplying and rearranging:
\[
\cot\frac{\alpha}{2}\cot\frac{\beta}{2}\cot\frac{\gamma}{2} - \cot\frac{\gamma}{2} = \cot\frac{\alpha}{2} + \cot\frac{\beta}{2}
\]
\[
\cot\frac{\alpha}{2} + \cot\frac{\beta}{2} + \cot\frac{\gamma}{2} = \cot\frac{\alpha}{2}\cot\frac{\beta}{2}\cot\frac{\gamma}{2}
\]
Answer: The conditional identity is proved.
Question 14: If \(\alpha+\beta+\gamma = 180^\circ\), show that \(\cot\alpha\cot\beta + \cot\beta\cot\gamma + \cot\gamma\cot\alpha = 1\)
Tan wali mashhoor identity \(\tan\alpha+\tan\beta+\tan\gamma = \tan\alpha\tan\beta\tan\gamma\) ko dono taraf product se divide karenge.
From \(\tan\alpha + \tan\beta + \tan\gamma = \tan\alpha\tan\beta\tan\gamma\), divide both sides by \(\tan\alpha\tan\beta\tan\gamma\):
\[
\frac{1}{\tan\beta\tan\gamma} + \frac{1}{\tan\alpha\tan\gamma} + \frac{1}{\tan\alpha\tan\beta} = 1
\]
\[
\cot\beta\cot\gamma + \cot\alpha\cot\gamma + \cot\alpha\cot\beta = 1
\]
Answer: The conditional identity is proved.
Question 15: Express the following in the form \(r\sin(\theta \pm \phi)\) where terminal sides of \(\theta\) and \(\phi\) are in the first quadrant: (i) \(24\sin\theta + 7\cos\theta\), (ii) \(12\sin\theta - 5\cos\theta\), (iii) \(\sin\theta - \cos\theta\), (iv) \(8\sin\theta - 6\cos\theta\), (v) \(\frac{1}{2}\sin\theta + \frac{\sqrt{3}}{2}\cos\theta\), (vi) \(13\sin\theta - 84\cos\theta\)
\(r = \sqrt{a^2+b^2}\) nikalenge aur \(\tan\phi = \frac{|b|}{|a|}\) se \(\phi\) (acute angle) maloom karenge. \(b\) positive ho to plus form, negative ho to minus form.
(i) \(r = \sqrt{24^2+7^2} = 25\), \(\phi = \tan^{-1}\frac{7}{24}\):
\[
24\sin\theta + 7\cos\theta = 25\sin(\theta + \phi), \quad \phi = \tan^{-1}\frac{7}{24}
\]
(ii) \(r = \sqrt{12^2+5^2} = 13\), \(\phi = \tan^{-1}\frac{5}{12}\):
\[
12\sin\theta - 5\cos\theta = 13\sin(\theta - \phi), \quad \phi = \tan^{-1}\frac{5}{12}
\]
(iii) \(r = \sqrt{1+1} = \sqrt{2}\), \(\phi = \tan^{-1}1 = 45^\circ\):
\[
\sin\theta - \cos\theta = \sqrt{2}\sin(\theta - 45^\circ)
\]
(iv) \(r = \sqrt{8^2+6^2} = 10\), \(\phi = \tan^{-1}\frac{6}{8} = \tan^{-1}\frac{3}{4}\):
\[
8\sin\theta - 6\cos\theta = 10\sin(\theta - \phi), \quad \phi = \tan^{-1}\frac{3}{4}
\]
(v) \(r = \sqrt{\frac{1}{4}+\frac{3}{4}} = 1\), \(\phi = \tan^{-1}\sqrt{3} = 60^\circ\):
\[
\frac{1}{2}\sin\theta + \frac{\sqrt{3}}{2}\cos\theta = \sin(\theta + 60^\circ)
\]
(vi) \(r = \sqrt{13^2+84^2} = 85\), \(\phi = \tan^{-1}\frac{84}{13}\):
\[
13\sin\theta - 84\cos\theta = 85\sin(\theta - \phi), \quad \phi = \tan^{-1}\frac{84}{13}
\]
Answer: (i) \(25\sin(\theta+\phi)\), (ii) \(13\sin(\theta-\phi)\), (iii) \(\sqrt{2}\sin(\theta-45^\circ)\), (iv) \(10\sin(\theta-\phi)\), (v) \(\sin(\theta+60^\circ)\), (vi) \(85\sin(\theta-\phi)\) with the indicated \(\phi\) values.
