{"id":31958,"date":"2023-01-03T16:56:52","date_gmt":"2023-01-03T11:26:52","guid":{"rendered":"https:\/\/apboardsolutions.in\/?p=31958"},"modified":"2023-01-03T16:56:52","modified_gmt":"2023-01-03T11:26:52","slug":"ap-inter-1st-year-maths-1a-solutions-chapter-6-ex-6c-in-telugu","status":"publish","type":"post","link":"https:\/\/apboardsolutions.in\/ap-inter-1st-year-maths-1a-solutions-chapter-6-ex-6c-in-telugu\/","title":{"rendered":"AP Inter 1st Year Maths 1A Solutions Chapter 6 \u0c24\u0c4d\u0c30\u0c3f\u0c15\u0c4b\u0c23\u0c2e\u0c3f\u0c24\u0c40\u0c2f \u0c28\u0c3f\u0c37\u0c4d\u0c2a\u0c24\u0c4d\u0c24\u0c41\u0c32\u0c41, \u0c2a\u0c30\u0c3f\u0c35\u0c30\u0c4d\u0c24\u0c28\u0c32\u0c41 Ex 6(c)"},"content":{"rendered":"

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions<\/a> Chapter 6 \u0c24\u0c4d\u0c30\u0c3f\u0c15\u0c4b\u0c23\u0c2e\u0c3f\u0c24\u0c40\u0c2f \u0c28\u0c3f\u0c37\u0c4d\u0c2a\u0c24\u0c4d\u0c24\u0c41\u0c32\u0c41, \u0c2a\u0c30\u0c3f\u0c35\u0c30\u0c4d\u0c24\u0c28\u0c32\u0c41 Exercise 6(c) will help students to clear their doubts quickly.<\/p>\n

AP Inter 1st Year Maths 1A Solutions Chapter 6 \u0c24\u0c4d\u0c30\u0c3f\u0c15\u0c4b\u0c23\u0c2e\u0c3f\u0c24\u0c40\u0c2f \u0c28\u0c3f\u0c37\u0c4d\u0c2a\u0c24\u0c4d\u0c24\u0c41\u0c32\u0c41, \u0c2a\u0c30\u0c3f\u0c35\u0c30\u0c4d\u0c24\u0c28\u0c32\u0c41 Exercise 6(c)<\/h2>\n

I.<\/span><\/p>\n

Question 1.
\n\u0c15\u0c3f\u0c02\u0c26\u0c3f \u0c35\u0c3e\u0c1f\u0c3f\u0c28\u0c3f \u0c38\u0c42\u0c15\u0c4d\u0c37\u0c4d\u0c2e\u0c40\u0c15\u0c30\u0c3f\u0c02\u0c1a\u0c02\u0c21\u0c3f.
\n(i) cos 100\u00b0 cos 40\u00b0 + sin 100\u00b0 sin 40\u00b0 [May ’12]
\nSolution:
\ncos 100\u00b0 cos 40\u00b0 + sin 100\u00b0 sin 40\u00b0 = cos(100\u00b0 – 40\u00b0)
\n= cos 60\u00b0
\n= \\(\\frac{1}{2}\\)<\/p>\n

(ii) \\(\\frac{\\cot 55 – \\cot 35}{\\cot 55+\\cot 35}\\)
\nSolution:
\n\\(\\frac{\\cot 55 – \\cot 35}{\\cot 55+\\cot 35}\\) = cot(55\u00b0 + 35\u00b0)
\n= cot (90\u00b0)
\n= 0<\/p>\n

(iii) \\(\\tan \\left[\\frac{\\pi}{4}+\\theta\\right] \\cdot \\tan \\left[\\frac{\\pi}{4}-\\theta\\right]\\)
\nSolution:
\n\\(\\left[\\frac{1+\\tan A}{1-\\tan A}\\right]\\left[\\frac{1-\\tan A}{1+\\tan A}\\right]=1\\)<\/p>\n

(iv) tan 75\u00b0 + cot 75\u00b0
\nSolution:
\ntan 75\u00b0 + cot 75\u00b0 = 2 + \u221a3 + 2 – \u221a3 = 4<\/p>\n

(v) sin 1140\u00b0 cos 390\u00b0 – cos 780\u00b0 sin 750\u00b0
\nSolution:
\nsin 1140\u00b0 cos 390\u00b0 – cos 780\u00b0 sin 750\u00b0
\n= sin(3 \u00d7 360\u00b0 + 60\u00b0) cos(360\u00b0 + 30\u00b0) – cos(2 \u00d7 360\u00b0 + 60\u00b0) sin(2 \u00d7 360\u00b0 + 30\u00b0)
\n= sin 60\u00b0 . cos 30\u00b0 – cos 60\u00b0 . sin 30\u00b0
\n= sin(60\u00b0 – 30\u00b0)
\n= sin 30\u00b0
\n= \\(\\frac{1}{2}\\)<\/p>\n

\"AP<\/p>\n

Question 2.
\n(i) \\(\\frac{\\sqrt{3} \\cos 25+\\sin 25}{2}\\) \u0c28\u0c41 sine \u0c15\u0c4b\u0c23\u0c02\u0c17\u0c3e \u0c30\u0c3e\u0c2f\u0c02\u0c21\u0c3f.
\nSolution:
\n\\(\\frac{\\sqrt{3} \\cos 25+\\sin 25}{2}\\)
\n= \\(\\frac{\\sqrt{3}}{2}\\) cos 25\u00b0 + \\(\\frac{1}{2}\\) sin 25\u00b0
\n= sin 60\u00b0 . cos 25\u00b0 + cos 60\u00b0 . sin 25\u00b0
\n= sin(60\u00b0 + 25\u00b0)
\n= sin 85\u00b0<\/p>\n

(ii) (cos \u03b8 – sin \u03b8) \u0c28\u0c41 cosine \u0c15\u0c4b\u0c23\u0c02\u0c17\u0c3e \u0c30\u0c3e\u0c2f\u0c02\u0c21\u0c3f.
\nSolution:
\n(cos \u03b8 – sin \u03b8)
\n\u221a2 \u0c28\u0c3f \u0c2d\u0c3e\u0c17\u0c3f\u0c02\u0c1a\u0c3f, \u0c17\u0c41\u0c23\u0c3f\u0c02\u0c1a\u0c17\u0c3e
\n\"AP<\/p>\n

(iii) sin(\u03b8 + \u03b1) = cos(\u03b8 + \u03b1) \u0c05\u0c2f\u0c3f\u0c24\u0c47, tan \u03b8 \u0c28\u0c41 tan \u03b1 \u0c2a\u0c26\u0c3e\u0c32\u0c32\u0c4b \u0c30\u0c3e\u0c2f\u0c02\u0c21\u0c3f.
\nSolution:
\ntan \u03b8 in term of tan \u03b1, if sin(\u03b8 + \u03b1) = cos(\u03b8 + \u03b1)
\n\u0c07\u0c1a\u0c4d\u0c1a\u0c3f\u0c28\u0c26\u0c3f sin(\u03b8 + \u03b1) = cos(\u03b8 + \u03b1)
\nsin \u03b8 cos \u03b1 + cos \u03b8 sin \u03b1 = cos \u03b8 cos \u03b1 – sin \u03b8 sin \u03b1
\ncos \u03b8 cos \u03b1 \u0c24\u0c4b \u0c2d\u0c3e\u0c17\u0c3f\u0c02\u0c1a\u0c17\u0c3e
\n\\(\\frac{\\sin \\theta \\cos \\alpha}{\\cos \\theta \\cos \\alpha}+\\frac{\\cos \\theta \\sin \\alpha}{\\cos \\theta \\cos \\alpha}\\) = \\(\\frac{\\cos \\theta \\cos \\alpha}{\\cos \\theta \\cos \\alpha}-\\frac{\\sin \\theta \\sin \\alpha}{\\cos \\theta \\cos \\alpha}\\)
\n\u21d2 tan \u03b8 + tan \u03b1 = 1 – tan \u03b8 tan \u03b1
\n\u21d2 tan \u03b8 + tan \u03b8 tan \u03b1 = 1 – tan \u03b1
\n\u21d2 tan \u03b8 (1 + tan \u03b1) = 1 – tan \u03b1
\n\u21d2 tan \u03b8 = \\(\\frac{1-\\tan \\alpha}{1+\\tan \\alpha}\\)<\/p>\n

Question 3.
\n(i) tan \u03b8 = \\(\\frac{\\cos 11^{\\circ}+\\sin 11^{\\circ}}{\\cos 11^{\\circ}-\\sin 11^{\\circ}}\\), \u03b8 \u0c2e\u0c42\u0c21\u0c35 \u0c2a\u0c3e\u0c26\u0c02\u0c32\u0c4b \u0c32\u0c47\u0c28\u0c3f \u0c15\u0c4b\u0c23\u0c02 \u03b8 \u0c28\u0c41 \u0c15\u0c28\u0c41\u0c15\u0c4d\u0c15\u0c4b\u0c02\u0c21\u0c3f.
\nSolution:
\n\u0c07\u0c1a\u0c4d\u0c1a\u0c3f\u0c28\u0c26\u0c3f tan \u03b8 = \\(\\frac{\\cos 11^{\\circ}+\\sin 11^{\\circ}}{\\cos 11^{\\circ}-\\sin 11^{\\circ}}\\)
\n= \\(\\frac{1+\\tan 11^{\\circ}}{1-\\tan 11^{\\circ}}\\)
\n= tan(45\u00b0 + 11\u00b0)
\n= tan(56\u00b0)
\n= tan(180\u00b0 + 56\u00b0)
\n= tan 236\u00b0
\n\u03b8 = 236\u00b0<\/p>\n

(ii) 0\u00b0 < A, B < 90\u00b0, \u0c05\u0c2f\u0c3f\u0c24\u0c47 cos A = \\(\\frac{5}{13}\\), sin B = \\(\\frac{4}{5}\\), \u0c05\u0c2f\u0c3f\u0c24\u0c47 sin(A – B) \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c15\u0c4d\u0c15\u0c4b\u0c02\u0c21\u0c3f.
\nSolution:
\n\"AP<\/p>\n

(iii) tan 20\u00b0 + tan 40\u00b0 + \u221a3 tan 20\u00b0 tan 40\u00b0 \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c15\u0c4d\u0c15\u0c4b\u0c02\u0c21\u0c3f.
\nSolution:
\nconsider 20\u00b0 + 40\u00b0 = 60\u00b0
\ntan(20\u00b0 + 40\u00b0) = tan 60\u00b0
\n\\(\\frac{\\tan 20^{\\circ}+\\tan 40^{\\circ}}{1-\\tan 20^{\\circ} \\tan 40^{\\circ}}\\) = \u221a3
\ntan 20\u00b0 + tan 40\u00b0 = \u221a3 – \u221a3 tan 20\u00b0 tan 40\u00b0
\ntan 20\u00b0 + tan 40\u00b0 + \u221a3 tan 20\u00b0 tan 40\u00b0 = \u221a3<\/p>\n

(iv) tan 56\u00b0 – tan 11\u00b0 – tan 56\u00b0 tan 11\u00b0 \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c15\u0c4d\u0c15\u0c4b\u0c02\u0c21\u0c3f.
\nSolution:
\nconsider 56\u00b0 – 11\u00b0 = 45\u00b0
\ntan(56\u00b0 – 11\u00b0) = tan 45\u00b0
\n\\(\\frac{\\tan 56^{\\circ}-\\tan 11^{\\circ}}{1+\\tan 56^{\\circ} \\tan 11^{\\circ}}\\) = 1
\ntan 56\u00b0 – tan 11\u00b0 = 1 + tan 56\u00b0 tan 11\u00b0
\ntan 56\u00b0 – tan 11\u00b0 – tan 56\u00b0 tan 11\u00b0 = 1<\/p>\n

(v) cos A, cos B, cos C \u0c32\u0c32\u0c4b \u0c0f \u0c12\u0c15\u0c4d\u0c15\u0c1f\u0c40 \u0c38\u0c41\u0c28\u0c4d\u0c28\u0c3e \u0c15\u0c3e\u0c15\u0c2a\u0c4b\u0c24\u0c47, \\(\\sum \\frac{\\sin (A+B) \\sin (A-B)}{\\cos ^2 A \\cos ^2 B}\\) \u0c28\u0c41 \u0c17\u0c23\u0c3f\u0c02\u0c1a\u0c02\u0c21\u0c3f.
\nSolution:
\n\\(\\sum \\frac{\\sin (A+B) \\sin (A-B)}{\\cos ^2 A \\cos ^2 B}\\)
\n\"AP<\/p>\n

(vi) sin A, sin B, sin C \u0c32\u0c32\u0c4b \u0c0f \u0c12\u0c15\u0c4d\u0c15\u0c1f\u0c40 \u0c38\u0c41\u0c28\u0c4d\u0c28\u0c3e \u0c15\u0c3e\u0c15\u0c2a\u0c4b\u0c24\u0c47, \\(\\sum \\frac{\\sin (C-A)}{\\sin C \\sin A}\\) \u0c28\u0c41 \u0c17\u0c23\u0c3f\u0c02\u0c1a\u0c02\u0c21\u0c3f.
\nSolution:
\n\"AP<\/p>\n

\"AP<\/p>\n

Question 4.
\n\u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c35\u0c3e\u0c1f\u0c3f\u0c28\u0c3f \u0c28\u0c3f\u0c30\u0c42\u0c2a\u0c3f\u0c02\u0c1a\u0c02\u0c21\u0c3f.
\n(i) cos 35\u00b0 + cos 85\u00b0 + cos 155\u00b0 = 0
\nSolution:
\ncos 35\u00b0 + cos 85\u00b0 + cos 155\u00b0
\n= -cos 85\u00b0 + 2 cos(\\(\\frac{35+155}{2}\\)) cos(\\(\\frac{35-155}{2}\\))
\n= -cos 85\u00b0 + 2 cos 85\u00b0 cos 60\u00b0
\n= -cos 85\u00b0 + 2 cos 85\u00b0 (\\(\\frac{1}{2}\\))
\n= -cos 85\u00b0 + cos 85\u00b0
\n= 0<\/p>\n

(ii) tan 72\u00b0 = tan 18\u00b0 + 2 tan 54\u00b0
\nSolution:
\ncot A – tan A = \\(\\frac{1}{\\tan A}\\) – tan A
\n= \\(\\frac{1-\\tan ^2 \\mathrm{~A}}{\\tan \\mathrm{A}}\\)
\n= \\(\\frac{2\\left(1-\\tan ^2 \\mathrm{~A}\\right)}{2 \\tan \\mathrm{A}}\\)
\n= \\(\\frac{2}{\\tan 2 A}\\)
\n= 2 cot 2A
\ncot A = tan A + 2 cot 2A
\nput A = 18\u00b0
\ncot 18\u00b0 = tan 18\u00b0 + 2 cot 36\u00b0
\ncot(90\u00b0 – 72\u00b0) = tan 18\u00b0 + 2 cot(90\u00b0 – 54\u00b0)
\ntan 72\u00b0 = tan 18\u00b0 + 2 tan 54\u00b0<\/p>\n

(iii) sin 750\u00b0 cos 480\u00b0 + cos 120\u00b0 cos 60\u00b0 = \\(\\frac{-1}{2}\\)
\nSolution:
\nsin 750\u00b0 = sin(2 \u00d7 360\u00b0 + 30\u00b0)
\n= sin 30\u00b0
\n= \\(\\frac{1}{2}\\)
\ncos 480\u00b0 = cos(360\u00b0 + 120\u00b0)
\n= cos 120\u00b0
\n= \\(\\frac{-1}{2}\\)
\nL.H.S. = sin 750\u00b0 cos 480\u00b0 + cos 120\u00b0 cos 60\u00b0
\n\"AP<\/p>\n

(iv) cos A + cos(\\(\\frac{4 \\pi}{3}\\) – A) + cos(\\(\\frac{4 \\pi}{3}\\) + A) = 0
\nSolution:
\ncos A + cos(\\(\\frac{4 \\pi}{3}\\) – A) + cos(\\(\\frac{4 \\pi}{3}\\) + A)
\n= cos A + 2 cos \\(\\frac{4 \\pi}{3}\\) cos A [\u2235 cos(A + B) + cos(A – B) = 2 cos A cos B]
\n= cos A + 2(\\(\\frac{-1}{2}\\)) cos A
\n= cos A – cos A
\n= 0<\/p>\n

(v) \\(\\cos ^2 \\theta+\\cos ^2\\left(\\frac{2 \\pi}{3}+\\theta\\right)+\\cos ^2\\left(\\frac{2 \\pi}{3}-\\theta\\right)=\\frac{3}{2}\\)
\nSolution:
\n\"AP
\n\"AP<\/p>\n

Question 5.
\n\u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c35\u0c3e\u0c1f\u0c3f\u0c28\u0c3f \u0c17\u0c23\u0c3f\u0c02\u0c1a\u0c02\u0c21\u0c3f.
\n(i) \\(\\sin ^2 82 \\frac{1}{2}^{\\circ}-\\sin ^2 22 \\frac{1^{\\circ}}{2}\\)
\nSolution:
\n\"AP<\/p>\n

(ii) \\(\\cos ^2 112 \\frac{1}{2}^{\\circ}-\\sin ^2 52 \\frac{1}{2}\\)
\nSolution:
\n\"AP<\/p>\n

(iii) \\(\\sin ^2\\left[\\frac{\\pi}{8}+\\frac{A}{2}\\right]-\\sin ^2\\left[\\frac{\\pi}{8}-\\frac{A}{2}\\right]\\)
\nSolution:
\n\\(\\sin ^2\\left[\\frac{\\pi}{8}+\\frac{A}{2}\\right]-\\sin ^2\\left[\\frac{\\pi}{8}-\\frac{A}{2}\\right]\\)
\n\"AP<\/p>\n

(iv) \\(\\cos ^2 52 \\frac{1}{2}^{\\circ}-\\sin ^2 22 \\frac{1}{2}^{\\circ}\\)
\nSolution:
\n\"AP<\/p>\n

\"AP<\/p>\n

Question 6.
\n\u0c15\u0c3f\u0c02\u0c26\u0c3f \u0c35\u0c3e\u0c1f\u0c3f\u0c15\u0c3f \u0c15\u0c28\u0c3f\u0c37\u0c4d\u0c20, \u0c17\u0c30\u0c3f\u0c37\u0c4d\u0c20 \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c32\u0c41 \u0c15\u0c28\u0c41\u0c15\u0c4d\u0c15\u0c4b\u0c02\u0c21\u0c3f.
\n(i) 3 cos x + 4 sin x
\nSolution:
\na = 4, b = 3, c = 0
\n\u0c15\u0c28\u0c3f\u0c37\u0c4d\u0c1f \u0c35\u0c3f\u0c32\u0c41\u0c35 = \\(c-\\sqrt{a^2+b^2}=\\sqrt{16+9}\\) = -5
\n\u0c17\u0c30\u0c3f\u0c37\u0c4d\u0c20 \u0c35\u0c3f\u0c32\u0c41\u0c35 = \\(c+\\sqrt{a^2+b^2}=\\sqrt{16+9}\\) = 5<\/p>\n

(ii) sin 2x – cos 2x
\nSolution:
\na = 1, b = -1, c = 0
\n\u0c15\u0c28\u0c3f\u0c37\u0c4d\u0c1f \u0c35\u0c3f\u0c32\u0c41\u0c35 = \\(c-\\sqrt{a^2+b^2}=-\\sqrt{1+1}\\) = -\u221a2
\n\u0c17\u0c30\u0c3f\u0c37\u0c4d\u0c20 \u0c35\u0c3f\u0c32\u0c41\u0c35 = \\(c+\\sqrt{a^2+b^2}=\\sqrt{1+1}\\) = \u221a2<\/p>\n

Question 7.
\n\u0c15\u0c3f\u0c02\u0c26\u0c3f \u0c35\u0c3e\u0c1f\u0c3f\u0c15\u0c3f \u0c35\u0c4d\u0c2f\u0c3e\u0c2a\u0c4d\u0c24\u0c3f \u0c15\u0c28\u0c41\u0c15\u0c4d\u0c15\u0c4b\u0c02\u0c21\u0c3f.
\n(i) 7 cos x – 24 sin x + 5
\nSolution:
\na = 24, b = 7, c = 5
\n\"AP<\/p>\n

(ii) 13 cos x + 3\u221a3 sin x – 4
\nSolution:
\na = 3\u221a3, b = 13, c = -4
\n\"AP<\/p>\n

II.<\/span><\/p>\n

Question 1.
\n(i) \\(\\frac{\\pi}{2}\\) < \u03b1 < \u03c0, 0 < \u03b2 < \\(\\frac{\\pi}{2}\\), cos \u03b1 = \\(\\frac{-3}{5}\\), sin \u03b2 = \\(\\frac{7}{25}\\) \u0c05\u0c2f\u0c3f\u0c24\u0c47 tan(\u03b1 + \u03b2), sin(\u03b1 + \u03b2) \u0c32 \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c32\u0c41 \u0c15\u0c28\u0c41\u0c15\u0c4d\u0c15\u0c4b\u0c02\u0c21\u0c3f.
\nSolution:
\n\"AP<\/p>\n

(ii) 0 < A < B < \\(\\frac{\\pi}{4}\\), sin(A + B) = \\(\\frac{24}{25}\\), cos(A – B) = \\(\\frac{4}{5}\\) \u0c05\u0c2f\u0c3f\u0c24\u0c47 tan 2A \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c15\u0c4d\u0c15\u0c4b\u0c02\u0c21\u0c3f. [(T.S) Mar. ’15]
\nSolution:
\n\"AP
\n\"AP<\/p>\n

(iii) A + B, A \u0c32\u0c41 \u0c32\u0c18\u0c41 \u0c15\u0c4b\u0c23\u0c3e\u0c32\u0c41 \u0c05\u0c35\u0c41\u0c24\u0c42 sin(A + B) = \\(\\frac{24}{25}\\), tan A = \\(\\frac{3}{4}\\) \u0c05\u0c2f\u0c3f\u0c24\u0c47, cos B \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c15\u0c4d\u0c15\u0c4b\u0c02\u0c21\u0c3f.
\nSolution:
\n\"AP<\/p>\n

(iv) tan \u03b1 – tan \u03b2 = m, cot \u03b1 – cot \u03b2 = n \u0c05\u0c2f\u0c3f\u0c24\u0c47, cot(\u03b1 – \u03b2) = \\(\\frac{1}{m}-\\frac{1}{n}\\) \u0c05\u0c28\u0c3f \u0c1a\u0c42\u0c2a\u0c02\u0c21\u0c3f.
\nSolution:
\ntan \u03b1 – tan \u03b2 = m
\n\u21d2 \\(\\frac{1}{\\cot \\alpha}-\\frac{1}{\\cot \\beta}\\) = m
\n\"AP<\/p>\n

(v) \u03b1, \u03b2 \u0c32\u0c41 \u0c2a\u0c4d\u0c30\u0c25\u0c2e \u0c2a\u0c3e\u0c26\u0c02\u0c32\u0c4b\u0c28\u0c3f \u0c15\u0c4b\u0c23\u0c3e\u0c32\u0c41, tan(\u03b1 – \u03b2) = \\(\\frac{7}{24}\\), tan \u03b1 = \\(\\frac{4}{3}\\) \u0c05\u0c2f\u0c3f\u0c24\u0c47, \u03b1 + \u03b2 = \\(\\frac{\\pi}{2}\\) \u0c05\u0c28\u0c3f \u0c1a\u0c42\u0c2a\u0c02\u0c21\u0c3f.
\nSolution:
\n\"AP<\/p>\n

\"AP<\/p>\n

Question 2.
\n(i) sin(A + B – C) \u0c35\u0c3f\u0c38\u0c4d\u0c24\u0c30\u0c23\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c15\u0c4d\u0c15\u0c4b\u0c02\u0c21\u0c3f.
\nSolution:
\nsin(A + B – C)
\n= sin[(A + B) – C]
\n= sin(A + B) . cos C – cos(A + B) sin C
\n= (sin A cos B + cos A sin B) cos C – (cos A cos B – sin A sin B) sin C
\n= sin A cos B cos C + cos A sin B cos C – cos A cos B sin C + sin A sin B sin C<\/p>\n

(ii) cos(A – B – C) \u0c35\u0c3f\u0c38\u0c4d\u0c24\u0c30\u0c23\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c15\u0c4d\u0c15\u0c4b\u0c02\u0c21\u0c3f.
\nSolution:
\ncos(A – B – C)
\n= cos{(A – B) – C}
\n= cos(A – B) cos C + sin(A – B) sin C
\n= (cos A cos B + sin A sin B) cos C + (sin A cos B – cos A sin B) sin C
\n= cos A cos B cos C + sin A sin B cos C + sin A cos B sin C – cos A sin B sin C<\/p>\n

(iii) \u2206ABC \u0c32\u0c4b A \u0c17\u0c41\u0c30\u0c41 \u0c15\u0c4b\u0c23\u0c02, sin A = \\(\\frac{3}{5}\\), sin B = \\(\\frac{5}{13}\\) \u0c05\u0c2f\u0c3f\u0c24\u0c47, sin C = \\(\\frac{16}{65}\\) \u0c05\u0c28\u0c3f \u0c1a\u0c42\u0c2a\u0c02\u0c21\u0c3f.
\nSolution:
\n\u0c07\u0c1a\u0c4d\u0c1a\u0c3f\u0c28\u0c26\u0c3f sin A = \\(\\frac{3}{5}\\)
\ncos2<\/sup>A = 1 – sin2<\/sup>A
\n= 1 – \\(\\frac{9}{25}\\)
\n= \\(\\frac{16}{25}\\)
\ncos A = \u00b1\\(\\frac{4}{5}\\)
\nA \u0c17\u0c41\u0c30\u0c41 \u0c15\u0c4b\u0c23\u0c02 \u21d2 90\u00b0 < A < 180\u00b0
\ntan A in II quadrant \u21d2 cos A is negative
\n\u2234 cos A = \\(\\frac{-4}{5}\\)
\n\u0c07\u0c1a\u0c4d\u0c1a\u0c3f\u0c28\u0c26\u0c3f sin \u03b2 = \\(\\frac{5}{13}\\)
\ncos2<\/sup>\u03b2 = 1 – sin2<\/sup>\u03b2
\n= 1 – \\(\\frac{25}{169}\\)
\n= \\(\\frac{144}{169}\\)
\ncos \u03b2 = \u00b1\\(\\frac{1}{2}\\)
\n\u03b2 is acute \u21d2 cos b is possible
\nsin \u03b2 = \\(\\frac{12}{13}\\)
\nA + B + C = 180\u00b0
\nC = 180\u00b0 – (A + B)
\nsin C = sin(180\u00b0 – (A + B))
\n= sin(A + B)
\n= sin A cos B + cos A sin B
\n= \\(\\left(\\frac{3}{5}\\right)\\left(\\frac{12}{13}\\right)+\\left(\\frac{-4}{5}\\right)\\left(\\frac{5}{13}\\right)\\)
\n= \\(\\frac{16}{65}\\)
\n\u2234 sec C = \\(\\frac{16}{65}\\)<\/p>\n

(iv) \\(\\frac{\\sin (\\alpha+\\beta)}{\\sin (\\alpha-\\beta)}=\\frac{a+b}{a-b}\\) \u0c05\u0c2f\u0c3f\u0c24\u0c47, a tan \u03b2 = tan \u03b1 \u0c05\u0c28\u0c3f \u0c1a\u0c42\u0c2a\u0c02\u0c21\u0c3f.
\nSolution:
\n\"AP
\n\"AP<\/p>\n

III.<\/span><\/p>\n

Question 1.
\n(i) A – B = \\(\\frac{3 \\pi}{4}\\) \u0c05\u0c2f\u0c3f\u0c24\u0c47, (1 – tan A) (1 + tan B) = 2 \u0c05\u0c28\u0c3f \u0c1a\u0c42\u0c2a\u0c02\u0c21\u0c3f.
\nSolution:
\nA – B = \\(\\frac{3 \\pi}{4}\\)
\ntan(A – B) = tan \\(\\frac{3 \\pi}{4}\\)
\n\\(\\frac{\\tan A-\\tan B}{1+\\tan A \\tan B}\\) = -1
\ntan A – tan B = -1 – tan A tan B
\n1 = -tan A + tan B – tan A tan B
\n2 = 1 – tan A + tan B – tan A tan B
\n2 = (1 – tan A) – tan B(1 – tan A)
\n(1 – tan A) (1 – tan B) = 2<\/p>\n

(ii) A + B + C = \\(\\frac{\\pi}{2}\\), A, B, C \u0c32\u0c32\u0c4b \u0c0f \u0c12\u0c15\u0c4d\u0c15\u0c1f\u0c40 \\(\\frac{\\pi}{2}\\) \u0c15\u0c3f \u0c2c\u0c47\u0c38\u0c3f \u0c17\u0c41\u0c23\u0c3f\u0c1c\u0c02 \u0c15\u0c3e\u0c15\u0c2a\u0c4b\u0c24\u0c47
\n(a) cot A + cot B + cot C = cot A cot B cot C \u0c05\u0c28\u0c3f \u0c1a\u0c42\u0c2a\u0c02\u0c21\u0c3f.
\n(b) tan A tan B + tan B tan C + tan C tan A = 1 \u0c05\u0c28\u0c3f \u0c1a\u0c42\u0c2a\u0c3f, \u0c26\u0c3e\u0c28\u0c3f \u0c28\u0c41\u0c02\u0c1a\u0c3f \\(\\sum \\frac{\\cos (B+C)}{\\cos B \\cos C}\\) = 2 \u0c05\u0c28\u0c3f \u0c1a\u0c42\u0c2a\u0c02\u0c21\u0c3f.
\nSolution:
\n\"AP
\n\"AP<\/p>\n

\"AP<\/p>\n

Question 2.
\n(i) sin2<\/sup>\u03b1 + cos2<\/sup>(\u03b1 + \u03b2) + 2 sin \u03b1 sin \u03b2 cos(\u03b1 + \u03b2) \u0c05\u0c28\u0c47\u0c26\u0c3f \u03b1 \u0c2a\u0c48 \u0c06\u0c27\u0c3e\u0c30\u0c2a\u0c21\u0c26\u0c28\u0c3f \u0c1a\u0c42\u0c2a\u0c02\u0c21\u0c3f.
\nSolution:
\nsin2<\/sup>\u03b1 + cos2<\/sup>(\u03b1 + \u03b2) + 2 sin \u03b1 cos(\u03b1 + \u03b2)
\n= sin2<\/sup>\u03b1 + cos(\u03b1 + \u03b2) (cos(\u03b1 + \u03b2) + 2 sin \u03b1 sin \u03b2)
\n= sin2<\/sup>\u03b1 + cos(\u03b1 + \u03b2) (cos \u03b1 cos \u03b2 – sin \u03b1 sin \u03b2 + 2 sin \u03b1 sin \u03b2)
\n= sin2<\/sup>\u03b1 + cos(\u03b1 + \u03b2) (cos \u03b1 cos \u03b2 + sin \u03b1 sin \u03b2)
\n= sin2<\/sup>\u03b1 + cos(\u03b1 + \u03b2) cos(\u03b1 – \u03b2)
\n= sin2<\/sup>\u03b1 + cos2<\/sup>\u03b2 – sin2<\/sup>\u03b1
\n= cos2<\/sup>\u03b2<\/p>\n

(ii) cos2<\/sup>(\u03b1 – \u03b2) + cos2<\/sup>\u03b2 – 2 cos(\u03b1 – \u03b2) cos \u03b1 cos \u03b2 \u0c05\u0c28\u0c47\u0c26\u0c3f \u03b2 \u0c2a\u0c48 \u0c06\u0c27\u0c3e\u0c30\u0c2a\u0c21\u0c26\u0c28\u0c3f \u0c1a\u0c42\u0c2a\u0c02\u0c21\u0c3f.
\nSolution:
\ncos2<\/sup>(\u03b1 – \u03b2) + cos2<\/sup>\u03b2 – 2 cos(\u03b1 – \u03b2) cos \u03b1 cos \u03b2
\n= cos2<\/sup>(\u03b1 – \u03b2) + cos2<\/sup>\u03b2 – cos(\u03b1 – \u03b2) [cos(\u03b1 + \u03b2) + cos(\u03b1 – \u03b2)]
\n= cos2<\/sup>(\u03b1 – \u03b2) + cos2<\/sup>\u03b2 – cos(\u03b1 – \u03b2) cos(\u03b1 + \u03b2) – cos2<\/sup>(\u03b1 – \u03b2)
\n= cos2<\/sup>\u03b2 – [cos2<\/sup>\u03b2 – sin2<\/sup>\u03b1]
\n= cos2<\/sup>\u03b2 – cos2<\/sup>\u03b2 + sin2<\/sup>\u03b1
\n= sin2<\/sup>\u03b1<\/p>\n","protected":false},"excerpt":{"rendered":"

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Chapter 6 \u0c24\u0c4d\u0c30\u0c3f\u0c15\u0c4b\u0c23\u0c2e\u0c3f\u0c24\u0c40\u0c2f \u0c28\u0c3f\u0c37\u0c4d\u0c2a\u0c24\u0c4d\u0c24\u0c41\u0c32\u0c41, \u0c2a\u0c30\u0c3f\u0c35\u0c30\u0c4d\u0c24\u0c28\u0c32\u0c41 Exercise 6(c) will help students to clear their doubts quickly. AP Inter 1st Year Maths 1A Solutions Chapter 6 \u0c24\u0c4d\u0c30\u0c3f\u0c15\u0c4b\u0c23\u0c2e\u0c3f\u0c24\u0c40\u0c2f \u0c28\u0c3f\u0c37\u0c4d\u0c2a\u0c24\u0c4d\u0c24\u0c41\u0c32\u0c41, \u0c2a\u0c30\u0c3f\u0c35\u0c30\u0c4d\u0c24\u0c28\u0c32\u0c41 Exercise 6(c) I. Question 1. \u0c15\u0c3f\u0c02\u0c26\u0c3f \u0c35\u0c3e\u0c1f\u0c3f\u0c28\u0c3f \u0c38\u0c42\u0c15\u0c4d\u0c37\u0c4d\u0c2e\u0c40\u0c15\u0c30\u0c3f\u0c02\u0c1a\u0c02\u0c21\u0c3f. (i) cos 100\u00b0 cos 40\u00b0 + sin 100\u00b0 sin 40\u00b0 [May … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[14],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/31958"}],"collection":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/comments?post=31958"}],"version-history":[{"count":2,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/31958\/revisions"}],"predecessor-version":[{"id":31986,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/31958\/revisions\/31986"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/media?parent=31958"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/categories?post=31958"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/tags?post=31958"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}