{"id":21005,"date":"2023-12-22T12:35:08","date_gmt":"2023-12-22T07:05:08","guid":{"rendered":"https:\/\/apboardsolutions.in\/?p=21005"},"modified":"2023-12-23T09:17:03","modified_gmt":"2023-12-23T03:47:03","slug":"ap-10th-class-maths-important-questions-chapter-11-in-telugu-medium","status":"publish","type":"post","link":"https:\/\/apboardsolutions.in\/ap-10th-class-maths-important-questions-chapter-11-in-telugu-medium\/","title":{"rendered":"AP 10th Class Maths Important Questions Chapter 11 \u0c24\u0c4d\u0c30\u0c3f\u0c15\u0c4b\u0c23\u0c2e\u0c3f\u0c24\u0c3f"},"content":{"rendered":"

These AP 10th Class Maths Chapter Wise Important Questions<\/a> 11th Lesson \u0c24\u0c4d\u0c30\u0c3f\u0c15\u0c4b\u0c23\u0c2e\u0c3f\u0c24\u0c3f will help students prepare well for the exams<\/p>\n

AP Board 10th Class Maths 11th Lesson Important Questions and Answers \u0c24\u0c4d\u0c30\u0c3f\u0c15\u0c4b\u0c23\u0c2e\u0c3f\u0c24\u0c3f<\/h2>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 1.
\nsin A = cos A \u0c05\u0c2f\u0c3f\u0c28 A \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c28\u0c41 \u0c21\u0c3f\u0c17\u0c4d\u0c30\u0c40\u0c32\u0c32\u0c4b \u0c24\u0c46\u0c32\u0c4d\u0c2a\u0c02\u0c21\u0c3f.
\n\u0c38\u0c3e\u0c27\u0c28.
\nsin A = cos A
\n\u21d2 sin A = sin (90 – A) (\u2235 A = 90 – A)
\n\u21d2 2A = 90
\n\u21d2 A = \\(\\frac{90}{2}\\) = 45\u00b0<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 2.
\n2 sin x = \u221a3 \u0c05\u0c2f\u0c3f\u0c28\u0c1a\u0c4b x \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c46\u0c02\u0c24?
\n\u0c38\u0c3e\u0c27\u0c28.
\n2 sin x = \u221a3
\n\u21d2 sin x = \\(\\frac{\\sqrt{3}}{2}\\)
\nsin x = sin 60\u00b0
\n\u21d2 x = 60\u00b0.<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 3.
\n\u0c17\u0c23\u0c3f\u0c02\u0c1a\u0c41\u0c2e\u0c41:
\n(i) cos 76\u00b0 – sin 14\u00b0
\n(ii) \\(\\frac{\\tan 73^{\\circ}}{\\cot 17^{\\circ}}\\)
\n\u0c38\u0c3e\u0c27\u0c28.
\n(i) cos 76\u00b0 – sin 14\u00b0 = cos (90 – 14) – sin 14
\n= sin 14\u00b0 – sin 14\u00b0 (\u2235 cos (90 – \u03b8) = sin \u03b8)
\n= 0
\n\u2234 cos 76\u00b0 – sin 14\u00b0 = 0 ………….(i)<\/p>\n

(i) \\(\\frac{\\tan 73^{\\circ}}{\\cot 17^{\\circ}}\\) = \\(\\frac{\\tan (90-17)}{\\cot 17}=\\frac{\\cot 17^{\\circ}}{\\cot 17^{\\circ}}\\)
\n= 1
\n\u2234 \\(\\frac{\\tan 73^{\\circ}}{\\cot 17^{\\circ}}\\) = 1<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 4.
\ntan2<\/sup> 45\u00b0 + cot2<\/sup> 30\u00b0 \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c15\u0c4d\u0c15\u0c4b\u0c02\u0c21\u0c3f.
\n\u0c38\u0c3e\u0c27\u0c28.
\ntan 45\u00b0 + cot2<\/sup> 30\u00b0
\n= (1)2<\/sup> + (\u221a3)2<\/sup>
\n= 1 + 3 = 4.<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 5.
\nsin2<\/sup> 30\u00b0 + cos2<\/sup> 60\u00b0 \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n\u0c38\u0c3e\u0c27\u0c28.
\nsin 30\u00b0 = \\(\\frac{1}{2}\\), cos 60\u00b0 = \\(\\frac{1}{2}\\)
\nsin2<\/sup> 30\u00b0 + cos2<\/sup> 60\u00b0 = (\\(\\frac{1}{2}\\))2<\/sup> + (\\(\\frac{1}{2}\\))2<\/sup>
\n= \\(\\frac{1}{4}+\\frac{1}{4}=\\frac{2}{4}=\\frac{1}{2}\\)
\n\u2234 sin2<\/sup> 30\u00b0 + cos2<\/sup> 60\u00b0 = \\(\\frac{1}{2}\\)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 6.
\nsin (A + B) = 1 \u0c2e\u0c30\u0c3f\u0c2f\u0c41 cos B = \\(\\frac{1}{2}\\) \u2220A, \u2220B\u0c32\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41. (0\u00b0< A + B \u2264 90\u00b0)
\n\u0c38\u0c3e\u0c27\u0c28.
\nsin (A + B) = 1
\nsin (A + B) = sin 90\u00b0
\nA + B = 90\u00b0
\ncos B = \\(\\frac{1}{2}\\)
\ncos B = cos 60\u00b0
\n\u2234 \u2220B = 60\u00b0
\n\u2234 \u2220A = 90\u00b0 – 60\u00b0 = 30\u00b0.<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 7.
\ncot2<\/sup> \u03b8 – \\(\\frac{1}{\\sin ^{2} \\theta}\\) \u0c28\u0c41 \u0c38\u0c42\u0c15\u0c4d\u0c37\u0c4d\u0c2e\u0c40\u0c15\u0c30\u0c3f\u0c02\u0c1a\u0c02\u0c21\u0c3f.
\n\u0c38\u0c3e\u0c27\u0c28.
\n\\(\\frac{\\cos ^{2} \\theta}{\\sin ^{2} \\theta}-\\frac{1}{\\sin ^{2} \\theta}=\\frac{\\cos ^{2} \\theta-1}{\\sin ^{2} \\theta}=\\frac{\\sin ^{2} \\theta}{\\sin ^{2} \\theta}\\) = 1<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 8.
\n‘\u03b8’ \u0c0f\u0c26\u0c47\u0c28\u0c3f \u0c05\u0c32\u0c4d\u0c2a\u0c15\u0c4b\u0c23\u0c2e\u0c48\u0c24\u0c47 sin \u03b8 = \\(\\frac{5}{3}\\); \u0c35\u0c4d\u0c2f\u0c35\u0c38\u0c4d\u0c25\u0c3f\u0c24\u0c2e\u0c17\u0c41\u0c28\u0c3e? .. \u0c15\u0c3e\u0c30\u0c23\u0c02 \u0c24\u0c46\u0c32\u0c2a\u0c02\u0c21\u0c3f.
\n\u0c38\u0c3e\u0c27\u0c28.
\n‘\u03b8’ \u0c05\u0c32\u0c4d\u0c2a\u0c15\u0c4b\u0c23\u0c2e\u0c41 = 0\u00b0 < \u03b8 < 90\u00b0
\n\u0c15\u0c28\u0c41\u0c15 sin 0\u00b0 = 0 \u0c2e\u0c30\u0c3f\u0c2f\u0c41 sin 90\u00b0 = 1
\n\u0c15\u0c28\u0c41\u0c15 0\u00b0 < \u03b8 < 90\u00b0\u0c15\u0c41 sin \u03b8 \u0c35\u0c3f\u0c32\u0c41\u0c35 0 \u0c2e\u0c30\u0c3f\u0c2f\u0c41 1\u0c32 \u0c2e\u0c27\u0c4d\u0c2f \u0c09\u0c02\u0c1f\u0c41\u0c02\u0c26\u0c3f.
\n\u0c15\u0c28\u0c41\u0c15 sin \u03b8 \u0c35\u0c3f\u0c32\u0c41\u0c35 1 \u0c15\u0c02\u0c1f\u0c47 \u0c0e\u0c15\u0c4d\u0c15\u0c41\u0c35 \u0c09\u0c02\u0c21\u0c26\u0c41.
\n\u2234 sin \u03b8 = \\(\\frac{5}{3}\\) \u0c35\u0c4d\u0c2f\u0c35\u0c38\u0c4d\u0c25\u0c24\u0c2e\u0c41 \u0c15\u0c3e\u0c26\u0c41.<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 9.
\nsin x = \\(\\frac{3}{4}\\) \u0c05\u0c2f\u0c3f\u0c24\u0c47 cosec x \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n\u0c38\u0c3e\u0c27\u0c28.
\nsin x = \\(\\frac{3}{4}\\) \u0c05\u0c2f\u0c3f\u0c24\u0c47
\ncosec x = \\(\\frac{1}{\\sin x}=\\frac{1}{\\frac{3}{4}}=\\frac{4}{3}\\)<\/p>\n

\"AP<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 10.
\nsin 90\u00b0, cos 90\u00b0, tan 90\u00b0, cot 90\u00b0, sec 90\u00b0 \u0c2e\u0c30\u0c3f\u0c2f\u0c41 cosec 90\u00b0 \u0c32\u0c32\u0c4b \u0c0f\u0c26\u0c3f( \u0c35\u0c3f) \u0c28\u0c3f\u0c30\u0c4d\u0c35\u0c39\u0c3f\u0c02\u0c1a\u0c2c\u0c21\u0c26\u0c41?
\n\u0c38\u0c3e\u0c27\u0c28.
\nsin 90\u00b0 = 1
\ncos 90\u00b0 = 0
\ntan 90\u00b0 = \u0c28\u0c3f\u0c30\u0c4d\u0c35\u0c39\u0c3f\u0c02\u0c1a\u0c2c\u0c21\u0c26\u0c41
\ncot 90\u00b0 = 0
\nsec 90\u00b0 = \u0c28\u0c3f\u0c30\u0c4d\u0c35\u0c39\u0c3f\u0c02\u0c1a\u0c2c\u0c21\u0c26\u0c41
\ncosec 90\u00b0 = 1
\n\u2234 tan 90\u00b0, sec 90\u00b0 \u0c32\u0c41 \u0c28\u0c3f\u0c30\u0c4d\u0c35\u0c39\u0c3f\u0c02\u0c1a\u0c2c\u0c21\u0c35\u0c41.<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 11.
\n(sec2<\/sup> x – 1) (cot2<\/sup> x) \u0c28\u0c41 \u0c38\u0c42\u0c15\u0c4d\u0c37\u0c4d\u0c2e \u0c30\u0c42\u0c2a\u0c02\u0c32\u0c4b \u0c24\u0c46\u0c32\u0c4d\u0c2a\u0c02\u0c21\u0c3f.
\n\u0c38\u0c3e\u0c27\u0c28.
\n(sec2<\/sup> x – 1) (cot2<\/sup> x) = (tan2<\/sup> x) (cot2<\/sup> x) [\u2235 sec2<\/sup>A – tan2<\/sup> A = 1]
\n= \\(\\frac{\\sin ^{2} x}{\\cos ^{2} x} \\cdot \\frac{\\cos ^{2} x}{\\sin ^{2} x}\\)[tan A = \\(\\frac{\\sin A}{\\cos A}\\), cot A = \\(\\frac{\\cos A}{\\sin A}\\)]
\n= 1.<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 12.
\ntan (A – B) = \\(\\frac{1}{\\sqrt{3}}\\) \u0c2e\u0c30\u0c3f\u0c2f\u0c41 sin A = \\(\\frac{\\sqrt{3}}{2}\\) \u0c05\u0c2f\u0c3f\u0c28\u0c1a\u0c4b \u2220B \u0c2e\u0c30\u0c3f\u0c2f\u0c41 cos B \u0c32 \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c32\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c02\u0c21\u0c3f (A, B < 90\u00b0)
\n\u0c38\u0c3e\u0c27\u0c28.
\ntan (A – B) = \\(\\frac{1}{\\sqrt{3}}\\)
\n\u2234 tan (A – B) = tan 30\u00b0
\n\u2234 A – B = 30\u00b0
\nsin A = \\(\\frac{\\sqrt{3}}{2}\\)
\nsin A = sin 60\u00b0
\n\u2234 A = 60\u00b0 \u0c2a\u0c4d\u0c30\u0c24\u0c3f\u0c15\u0c4d\u0c37\u0c47\u0c2a\u0c3f\u0c02\u0c1a\u0c3f\u0c28
\nA – B = 30\u00b0
\n\u21d2 60\u00b0 – B = 30\u00b0
\n\u21d2 B = 30\u00b0
\n\u21d2 cos B = cos 30\u00b0
\n\u21d2 cos B = \\(\\frac{\\sqrt{3}}{2}\\)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 13.
\ntan A = \\(\\frac{1}{\\sqrt{3}}\\) \u0c2e\u0c30\u0c3f\u0c2f\u0c41 tan B = \u221a3 \u0c05\u0c2f\u0c3f\u0c28 sin A . cos B + cos A . sin B \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41 (A, B < 90\u00b0)
\n\u0c38\u0c3e\u0c27\u0c28.
\ntan A = \\(\\frac{1}{\\sqrt{3}}\\) \u0c2e\u0c30\u0c3f\u0c2f\u0c41 tan B = \u221a3 (A, B < 90\u00b0)
\n\u2234 tan A = \\(\\frac{1}{\\sqrt{3}}\\) = tan 30\u00b0 (\u2235 A < 90\u00b0)
\n\u21d2 A = 30\u00b0 \u0c2e\u0c30\u0c3f\u0c2f\u0c41 tan B = \u221a3 = tan 60\u00b0 \u21d2 B = 60\u00b0
\n(\u2235 AB < 90\u00b0)
\n\u21d2 A = 30\u00b0, B = 60\u00b0 \u0c05\u0c2f\u0c3f\u0c28
\nsin A cos B + cos A sin B \u0c35\u0c3f\u0c32\u0c41\u0c35 = sin 30\u00b0 cos 60\u00b0 + cos 30\u00b0 sin 60\u00b0
\n\u2234 sin 30\u00b0 cos 60\u00b0 + cos 30\u00b0 sin 60\u00b0 = \\(\\frac{1}{2} \\cdot \\frac{1}{2}+\\frac{\\sqrt{3}}{2} \\cdot \\frac{\\sqrt{3}}{2}\\)
\n= \\(\\frac{1}{4}+\\frac{3}{4}=\\frac{4}{4}\\)
\n= 1<\/p>\n

IInd method :
\nsin A cos B + cos A sin. B = sin (A + B) \u0c28\u0c02\u0c26\u0c41
\nA = 30\u00b0, B = 60\u00b0 \u0c2a\u0c4d\u0c30\u0c24\u0c3f\u0c15\u0c4d\u0c37\u0c47\u0c2a\u0c3f\u0c02\u0c1a\u0c3f\u0c28
\nsin (A + B) = sin (30\u00b0 + 60\u00b0)
\n= sin 90\u00b0 = 1.<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 14.
\ntan2<\/sup> A – sin2<\/sup> A = tan2<\/sup> A . sin2<\/sup> A \u0c05\u0c28\u0c3f \u0c1a\u0c42\u0c2a\u0c02\u0c21\u0c3f.
\n\u0c38\u0c3e\u0c27\u0c28.
\ntan2<\/sup> A – sin2<\/sup> A = \\(\\frac{\\sin ^{2} A}{\\cos ^{2} A}\\) – sin2<\/sup> A
\n= sin2<\/sup> A (\\(\\frac{1}{\\cos ^{2} A}\\) – 1)
\n= sin2<\/sup> A (sec2<\/sup> A – 1)
\n= sin2<\/sup> A. tan2<\/sup> A<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 15.
\ncos A = \\(\\frac{7}{25}\\) \u0c05\u0c2f\u0c3f\u0c28 sin A \u0c2e\u0c30\u0c3f\u0c2f\u0c41 cosec A \u0c32\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c02\u0c21\u0c3f. \u0c28\u0c40\u0c35\u0c47\u0c2e\u0c3f \u0c17\u0c2e\u0c28\u0c3f\u0c02\u0c1a\u0c3f\u0c24\u0c3f\u0c35\u0c3f ?
\n\u0c38\u0c3e\u0c27\u0c28.<\/p>\n

\"AP<\/p>\n

\u2206ABC \u0c32\u0c02\u0c2c\u0c15\u0c4b\u0c23 \u0c24\u0c4d\u0c30\u0c3f\u0c2d\u0c41\u0c1c\u0c30\u0c40\u0c32\u0c4b
\ncos A = \\(\\frac{7}{25}=\\frac{\\mathrm{AB}}{\\mathrm{AC}}\\)
\nx2<\/sup> + 72<\/sup> = 252<\/sup>
\nx2<\/sup> = 252<\/sup> – 72<\/sup> = 576
\nx = 24
\nsin A = \\(\\frac{24}{25}\\) cosec A = \\(\\frac{25}{24}\\)
\ncosec A = \\(\\frac{1}{\\sin A}\\) \u0c05\u0c28\u0c3f \u0c2a\u0c30\u0c3f\u0c36\u0c40\u0c32\u0c3f\u0c02\u0c1a\u0c3f\u0c24\u0c3f\u0c28\u0c3f.<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 16.
\ntan 2A = cot (A – 18\u00b0), 2A \u0c12\u0c15 \u0c05\u0c32\u0c4d\u0c2a\u0c15\u0c4b\u0c23\u0c2e\u0c41 \u0c05\u0c2f\u0c3f\u0c24\u0c47 A \u0c2f\u0c4a\u0c15\u0c4d\u0c15 \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n\u0c38\u0c3e\u0c27\u0c28.
\ntan 2A = cot(A – 18\u00b0)
\n= cot[90 – (90 – (A – 18\u00b0)]]
\ntan 2A = tan [90 – (A – 18\u00b0)]
\n2A = 90 – (A – 18\u00b0)
\n= 90 – A + 18\u00b0
\n\u21d2 3A = 108\u00b0
\n\u2234 A = 36\u00b0<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 17.
\n4 tan \u03b8 = 3 \u0c05\u0c2f\u0c3f\u0c28 sec e \u0c2e\u0c30\u0c3f\u0c2f\u0c41 cosec 2\u0c32 \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c32\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n\u0c38\u0c3e\u0c27\u0c28.
\n4 tan \u03b8 = 3
\n\u21d2 tan \u03b8 = \\(\\frac{3}{4}\\)<\/p>\n

\"AP<\/p>\n

AC = \\(\\sqrt{\\mathrm{BC}^{2}+\\mathrm{AB}^{2}}\\)
\n= \\(\\sqrt{4^{2}+3^{2}}\\) = 5
\nsec \u03b8 = \\(\\frac{5}{4}\\); cosec \u03b8 = \\(\\frac{5}{3}\\).<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 18.
\ntan 2A = cot (A – 27), 2A \u0c05\u0c32\u0c4d\u0c2a\u0c15\u0c4b\u0c23\u0c2e\u0c48\u0c28 A \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c02\u0c21\u0c3f.
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u0c26\u0c24\u0c4d\u0c24\u0c3e\u0c02\u0c36\u0c02 \u0c2a\u0c4d\u0c30\u0c15\u0c3e\u0c30\u0c02, 2A \u0c05\u0c32\u0c4d\u0c2a\u0c15\u0c4b\u0c23\u0c2e\u0c41
\ntan 2A = cot (A – 27)
\n\u21d2 2A + A – 27 = 90
\n\u21d2 2A + A = 90 + 27 (\u2235 tan (90 – \u03b8) = cot \u03b8)
\n\u21d2 3A = 117
\n\u21d2 A = \\(\\frac{117}{3}\\) = 39
\n\u2234 A = 39\u00b0<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 19.
\n\\(\\sqrt{\\frac{1+\\sin A}{1-\\sin A}}\\) = sec A + tan A \u0c05\u0c28\u0c3f \u0c1a\u0c42\u0c2a\u0c02\u0c21\u0c3f.
\n\u0c38\u0c3e\u0c27\u0c28.
\n\\(\\sqrt{\\frac{1+\\sin A}{1-\\sin A}}\\)
\n(\u0c32\u0c35, \u0c39\u0c3e\u0c30\u0c3e\u0c32\u0c28\u0c41 \\(\\sqrt{1+\\sin A}\\) \u0c24\u0c4b \u0c17\u0c41\u0c23\u0c3f\u0c02\u0c1a\u0c3f\u0c28)<\/p>\n

\"AP<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 20.
\n(sin x – cos x)2<\/sup> + (sin x + cos x)2<\/sup> \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c02\u0c21\u0c3f.
\n\u0c38\u0c3e\u0c27\u0c28.
\n(sin x – cos x)2<\/sup> + (sin x + cos x)2<\/sup> = sin2<\/sup> x + cos2<\/sup> x – 2 sin x cos x + sin2<\/sup> x + cos2<\/sup> x + 2 sin x cos x
\n= 2(sin2<\/sup> x + cos2<\/sup> x) = 2(1) = 2<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 21.
\ncos A = \\(\\frac{12}{13}\\) \u0c05\u0c2f\u0c3f\u0c24\u0c47 sin A \u0c2e\u0c30\u0c3f\u0c2f\u0c41 tan A \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c02\u0c21\u0c3f.
\n\u0c38\u0c3e\u0c27\u0c28.<\/p>\n

\"AP<\/p>\n

AC2<\/sup> = AB2<\/sup> + BC2<\/sup>
\n132<\/sup> = 122<\/sup> + x2<\/sup>
\n169 = 144 + x2<\/sup>
\nx2<\/sup> = 25
\n\u21d2 x = 5
\nsin A = \\(\\frac{5}{13}\\)
\ntan A = \\(\\frac{\\sin A}{\\cos A}=\\frac{\\frac{5}{13}}{\\frac{12}{13}}=\\frac{5}{13} \\times \\frac{13}{12}=\\frac{5}{12}\\)
\n\u2234 sin A = \\(\\frac{5}{13}\\), tan A = \\(\\frac{5}{12}\\).<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 22.
\n4 sin2<\/sup> \u03b8 – 1 = 0 \u0c05\u0c2f\u0c3f\u0c28 (\u03b8 < 90) \u03b8 \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c3f cos2<\/sup> \u03b8 + tan2<\/sup> \u03b8 \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c02\u0c21\u0c3f
\n\u0c38\u0c3e\u0c27\u0c28.
\n4 sin2<\/sup> \u03b8 – 1 = 0
\n\u21d2 4 sin2<\/sup> \u03b8 = 1
\nsin2<\/sup> \u03b8 = \\(\\frac{1}{4}\\)
\n\u21d2 sin \u03b8 = \u00b1 \\(\\sqrt{\\frac{1}{4}}\\)
\n= \u00b1 \\(\\frac{1}{2}\\)
\n\u03b8 < 90\u00b0. \u0c05\u0c28\u0c3f \u0c07\u0c35\u0c4d\u0c35\u0c2c\u0c21\u0c3f\u0c02\u0c26\u0c3f
\n\u0c15\u0c3e\u0c2c\u0c1f\u0c4d\u0c1f\u0c3f sin \u03b8 = \\(\\frac{1}{2}\\)
\n\u2234 \u03b8 = 30\u00b0
\ncos \u03b8 = cos 30\u00b0 = \\(\\frac{\\sqrt{3}}{2}\\)
\ntan \u03b8 = tan 30\u00b0 = \\(\\frac{1}{\\sqrt{3}}\\)
\ncos2<\/sup> \u03b8 + tan2<\/sup> \u03b8 = \\(\\left(\\frac{\\sqrt{3}}{2}\\right)^{2}+\\left(\\frac{1}{\\sqrt{3}}\\right)^{2}\\)
\n= \\(\\frac{3}{4}+\\frac{1}{3}=\\frac{9+4}{12}=\\frac{13}{12}\\)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 23.
\n(sin \u03b8 – cosec \u03b8)2<\/sup> + (cos \u03b8 – sec \u03b8)2<\/sup> = cot \u03b8 + tan2<\/sup> \u03b8 – 1 \u0c05\u0c28\u0c3f \u0c28\u0c3f\u0c30\u0c42\u0c2a\u0c3f\u0c02\u0c1a\u0c02\u0c21\u0c3f.
\n\u0c38\u0c3e\u0c27\u0c28.
\n(sin \u03b8 – cosec \u03b8)2<\/sup> + (cos \u03b8 – sec \u03b8)2<\/sup> = sin2<\/sup> \u03b8 + cosec2<\/sup> \u03b8 – 2 sin \u03b8 . cosec \u03b8 + cos2<\/sup> \u03b8 + sec2<\/sup> \u03b8 – 2 cos \u03b8 \u00b7 sec \u03b8
\n= (sin2<\/sup> \u03b8 + cos2<\/sup> \u03b8) + cosec2<\/sup> \u03b8 + sec2<\/sup> \u03b8 – 2 – 2
\n= 1 + (1 + cot2<\/sup> \u03b8) + (1 + tan2<\/sup> \u03b8) – 2 – 2
\n= cot2<\/sup> \u03b8 + tan2<\/sup> \u03b8 + 3 – 4
\n= cot2<\/sup> \u03b8 + tan 2<\/sup> \u03b8 – 1.<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 24.
\ncosec \u03b8 + cot \u03b8 = p \u0c05\u0c2f\u0c3f\u0c24\u0c47 \\(\\frac{p^{2}+1}{p^{2}-1}\\) = sec \u03b8 \u0c05\u0c28\u0c3f \u0c1a\u0c42\u0c2a\u0c02\u0c21\u0c3f.
\n\u0c38\u0c3e\u0c27\u0c28.
\ncosec \u03b8 + cot \u03b8 = p.
\n\u2234 cosec \u03b8 – cot \u03b8 = \\(\\frac{1}{p}\\)
\ncosec \u03b8 + cot \u03b8 = p
\n2 cosec \u03b8 = p + \\(\\frac{1}{p}\\)
\ncosec \u03b8 – cot \u03b8 = \\(\\frac{1}{p}\\)
\n2 cot \u03b8 = p – \\(\\frac{1}{p}\\)
\n= \\(\\frac{p^{2}-1}{p}\\)
\n\\(\\frac{\\frac{p^{2}+1}{p}}{\\frac{p^{2}-1}{p}}=\\frac{2 \\cosec \\theta}{2 \\cot \\theta}\\)<\/p>\n

\\(\\frac{\\mathrm{p}^{2}+1}{\\mathrm{p}^{2}-1}=\\frac{\\frac{1}{\\sin \\theta}}{\\frac{\\cos \\theta}{\\sin \\theta}}=\\frac{1}{\\cos \\theta}\\) = sec \u03b8.<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 25.
\nsec \u03b8 + tan \u03b8 = p. \u0c05\u0c2f\u0c3f\u0c28 sin \u03b8 = \\(\\frac{p^{2}-1}{p^{2}+1}\\) \u0c05\u0c28\u0c3f \u0c28\u0c3f\u0c30\u0c42\u0c2a\u0c3f\u0c02\u0c1a\u0c02\u0c21\u0c3f.
\n\u0c38\u0c3e\u0c27\u0c28.
\nsec \u03b8 + tan \u03b8 = p
\nsec2<\/sup> \u03b8 – tan2<\/sup> \u03b8 = 1
\n(sec \u03b8 + tan \u03b8) (sec \u03b8 – tan \u03b8) = 1
\np. (sec \u03b8 – tan \u03b8) = 1<\/p>\n

\"AP<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 26.
\ncot \u03b8 = \\(\\frac{7}{8}\\) \u0c05\u0c2f\u0c3f\u0c28
\n(i) \\(\\frac{(1+\\sin \\theta)(1-\\sin \\theta)}{(1+\\cos \\theta)(1-\\cos \\theta)}\\)
\n(ii) \\(\\frac{1+\\cos \\theta}{\\sin \\theta}\\) \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c32\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c66\u0c21\u0c3f.
\n\u0c38\u0c3e\u0c27\u0c28.
\n(i) \\(\\frac{(1+\\sin \\theta)(1-\\sin \\theta)}{(1+\\cos \\theta)(1-\\cos \\theta)}\\) = \\(\\frac{1-\\sin ^{2} \\theta}{1-\\cos ^{2} \\theta}=\\frac{\\cos ^{2} \\theta}{\\sin ^{2} \\theta}\\)
\n= cot2<\/sup> \u03b8.
\n= \\(\\left(\\frac{7}{8}\\right)^{2}=\\frac{49}{64}\\) \u2192 (1)
\n(\u2235 sin2<\/sup> \u03b8 + cos2<\/sup> \u03b8 = 1)<\/p>\n

(ii) \\(\\frac{1+\\cos \\theta}{\\sin \\theta}\\) = \\(\\frac{1}{\\sin \\theta}+\\frac{\\cos \\theta}{\\sin \\theta}\\)
\n= cosec \u03b8 + cot \u03b8
\ncot \u03b8 = \\(\\frac{7}{8}\\) \u0c15\u0c3e\u0c35\u0c41\u0c28, (1 + cot2<\/sup> \u03b8) = cosec2<\/sup> \u03b8
\n\u21d2 1 + (\\(\\frac{7}{8}\\))2<\/sup> = cosec2<\/sup> \u03b8
\n\u21d2 1 + \\(\\frac{49}{64}\\) = \\(\\frac{64+49}{64}=\\frac{113}{64}\\)
\n\u2234 cosec \u03b8 = \\(\\sqrt{\\frac{113}{64}}\\)
\n\u2234 \\(\\frac{1+\\cos \\theta}{\\sin \\theta}\\) = cosec \u03b8 + cot \u03b8
\n= \\(\\frac{\\sqrt{113}}{8}+\\frac{7}{8}=\\frac{7+\\sqrt{113}}{8}\\).<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 27.
\nsec2<\/sup> \u03b8 + cosec2<\/sup> \u03b8 = sec2<\/sup> \u03b8 . cosec2<\/sup> \u03b8 .
\n\u0c38\u0c3e\u0c27\u0c28.
\nsec2<\/sup> \u03b8 + cosec2<\/sup> \u03b8 = \\(\\frac{1}{\\cos ^{2} \\theta}+\\frac{1}{\\sin ^{2} \\theta}\\)
\n= \\(\\frac{\\sin ^{2} \\theta+\\cos ^{2} \\theta}{\\sin ^{2} \\theta \\cdot \\cos ^{2} \\theta}\\)<\/p>\n

= \\(\\frac{1}{\\sin ^{2} \\theta \\cdot \\cos ^{2} \\theta}\\)<\/p>\n

= \\(\\frac{1}{\\sin ^{2} \\theta} \\cdot \\frac{1}{\\cos ^{2} \\theta}\\)<\/p>\n

= cosec2<\/sup> \u03b8 . sec2<\/sup> \u03b8<\/p>\n","protected":false},"excerpt":{"rendered":"

These AP 10th Class Maths Chapter Wise Important Questions 11th Lesson \u0c24\u0c4d\u0c30\u0c3f\u0c15\u0c4b\u0c23\u0c2e\u0c3f\u0c24\u0c3f will help students prepare well for the exams AP Board 10th Class Maths 11th Lesson Important Questions and Answers \u0c24\u0c4d\u0c30\u0c3f\u0c15\u0c4b\u0c23\u0c2e\u0c3f\u0c24\u0c3f \u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 1. sin A = cos A \u0c05\u0c2f\u0c3f\u0c28 A \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c28\u0c41 \u0c21\u0c3f\u0c17\u0c4d\u0c30\u0c40\u0c32\u0c32\u0c4b \u0c24\u0c46\u0c32\u0c4d\u0c2a\u0c02\u0c21\u0c3f. \u0c38\u0c3e\u0c27\u0c28. sin A = cos A \u21d2 sin A = sin … 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":[13],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/21005"}],"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=21005"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/21005\/revisions"}],"predecessor-version":[{"id":21006,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/21005\/revisions\/21006"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/media?parent=21005"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/categories?post=21005"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/tags?post=21005"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}