{"id":21326,"date":"2023-12-27T12:46:49","date_gmt":"2023-12-27T07:16:49","guid":{"rendered":"https:\/\/apboardsolutions.in\/?p=21326"},"modified":"2023-12-28T09:38:25","modified_gmt":"2023-12-28T04:08:25","slug":"ap-board-10th-class-maths-solutions-chapter-11-exercise-11-4-in-telugu","status":"publish","type":"post","link":"https:\/\/apboardsolutions.in\/ap-board-10th-class-maths-solutions-chapter-11-exercise-11-4-in-telugu\/","title":{"rendered":"AP Board 10th Class Maths Solutions Chapter 11 \u0c24\u0c4d\u0c30\u0c3f\u0c15\u0c4b\u0c23\u0c2e\u0c3f\u0c24\u0c3f Exercise 11.4"},"content":{"rendered":"

SCERT AP 10th Class Maths Textbook Solutions<\/a> Chapter 11 \u0c24\u0c4d\u0c30\u0c3f\u0c15\u0c4b\u0c23\u0c2e\u0c3f\u0c24\u0c3f Exercise 11.4 Textbook Exercise Questions and Answers.<\/p>\n

AP State Syllabus 10th Class Maths Solutions 11th Lesson \u0c24\u0c4d\u0c30\u0c3f\u0c15\u0c4b\u0c23\u0c2e\u0c3f\u0c24\u0c3f Exercise 11.4<\/h2>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 1.
\n\u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c35\u0c3e\u0c1f\u0c3f\u0c28\u0c3f \u0c38\u0c42\u0c15\u0c4d\u0c37\u0c4d\u0c2e\u0c3f\u0c15\u0c30\u0c3f\u0c02\u0c1a\u0c02\u0c21\u0c3f:
\n(i) (1 + tan \u03b8 + sec \u03b8) (1 + cot \u03b8 – cosec \u03b8)
\n\u0c38\u0c3e\u0c27\u0c28.
\n(1 + tan \u03b8 + sec \u03b8) (1 + cot \u03b8 – cosec \u03b8)
\n= \\(\\left(\\frac{1}{1}+\\frac{\\sin \\theta}{\\cos \\theta}+\\frac{1}{\\cos \\theta}\\right)\\left(\\frac{1}{1}+\\frac{\\cos \\theta}{\\sin \\theta}-\\frac{1}{\\sin \\theta}\\right)\\)
\n= \\(\\left(\\frac{(\\cos \\theta+\\sin \\theta)+1}{\\cos \\theta}\\right)\\left(\\frac{(\\sin \\theta+\\cos \\theta)-1}{\\sin \\theta}\\right)\\)
\n[\u2235 (a + b) (a – b) = a2<\/sup> – b2<\/sup>] cose + sin\u2019e +2 sino cose – 1 :: :coso. sine
\n= \\(\\frac{(\\cos \\theta+\\sin \\theta)^{2}-(1)^{2}}{\\cos \\theta \\cdot \\sin \\theta}\\)
\n= \\(\\frac{\\cos ^{2} \\theta+\\sin ^{2} \\theta+2 \\sin \\theta \\cos \\theta-1}{\\cos \\theta \\cdot \\sin \\theta}\\)
\n= \\(\\frac{1+2 \\sin \\theta \\cdot \\cos \\theta-1}{\\cos \\theta \\cdot \\sin \\theta}\\) [\u2235 cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1]
\n= \\(\\frac{2 \\sin \\theta \\cdot \\cos \\theta}{\\cos \\theta \\cdot \\sin \\theta}\\) = 2.<\/p>\n

\"AP<\/p>\n

(ii) (sin \u03b8 + cos \u03b8)2<\/sup> + (sin \u03b8 – cos \u03b8)2<\/sup>
\n\u0c38\u0c3e\u0c27\u0c28.
\n(sin \u03b8 + cos \u03b8)2<\/sup> + (sin \u03b8 – cos \u03b8)2<\/sup> = (sin2<\/sup> \u03b8 + cos2<\/sup> \u03b8 + 2 sin \u03b8 cos \u03b8) +
\n(sin2<\/sup> \u03b8 + cos2<\/sup> \u03b8 – 2 sin \u03b8 cos \u03b8)
\n[\u2235 (a + b)2<\/sup> = a2<\/sup> + b2<\/sup> + 2ab
\n(a – b)2<\/sup> = a2<\/sup> + b2<\/sup> – 2ab]
\n= 1 + 2 sin \u03b8 cos \u03b8 + 1 – 2 sin \u03b8 cos \u03b8 [\u2235 sin2<\/sup> \u03b8 + cos2<\/sup> \u03b8 = 1]
\n= 1 + 1 = 2<\/p>\n

(iii) (sec2<\/sup> \u03b8 – 1) (cosec2<\/sup> \u03b8 – 1)
\n\u0c38\u0c3e\u0c27\u0c28.
\n(sec2<\/sup> \u03b8 – 1) (cosec2<\/sup> \u03b8 – 1) = tan2<\/sup> \u03b8 \u00d7 cot2<\/sup> \u03b8
\n[\u2235 sec2<\/sup> \u03b8 – tan2<\/sup> \u03b8 = 1
\ncosec2<\/sup> \u03b8 – cot2<\/sup> \u03b8 = 1]
\n= tan2<\/sup> \u03b8 \u00d7 \\(\\frac{1}{\\tan ^{2} \\theta}\\) = 1<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 2.
\n(cosec \u03b8 – cot \u03b8)2<\/sup> = \\(\\frac{1-\\cos \\theta}{1+\\cos \\theta}\\) \u0c05\u0c28\u0c3f \u0c1a\u0c42\u0c2a\u0c3f\u0c02\u0c1a\u0c02\u0c21\u0c3f
\n\u0c38\u0c3e\u0c27\u0c28.
\nL.H.S. = (cosec \u03b8 – cot \u03b8)2<\/sup>
\n= \\(\\left(\\frac{1}{\\sin \\theta}-\\frac{\\cos \\theta}{\\sin \\theta}\\right)^{2}\\)
\n= \\(\\frac{(1-\\cos \\theta)^{2}}{\\sin ^{2} \\theta}\\)
\n= \\(\\frac{(1-\\cos \\theta)^{2}}{1^{2}-\\cos ^{2} \\theta}\\)
\n[\u2235 sin2<\/sup> \u03b8 = 1 – cos2<\/sup> \u03b8]
\n= \\(\\)
\n[\u2235 a2<\/sup> – b2<\/sup> = (a + b) (a – b)]
\n= \\(\\frac{1-\\cos \\theta}{1+\\cos \\theta}\\) = R.H.S.<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 3.
\n\\(\\sqrt{\\frac{1+\\sin A}{1-\\sin A}}\\) = sec A + tan A \u0c1a\u0c42\u0c2a\u0c02\u0c21\u0c3f.
\n\u0c38\u0c3e\u0c27\u0c28.
\nL.H.S = \\(\\sqrt{\\frac{1+\\sin A}{1-\\sin A}}\\)
\n\u0c32\u0c35\u0c39\u0c3e\u0c30\u0c3e\u0c32\u0c28\u0c41 1 – sin A \u0c24\u0c4b \u0c17\u0c23\u0c3f\u0c02\u0c1a\u0c17\u0c3e
\n= \\(\\sqrt{\\frac{1+\\sin A}{1-\\sin A} \\times \\frac{1+\\sin A}{1+\\sin A}}\\)
\n= \\(\\sqrt{\\frac{(1+\\sin A)^{2}}{1-\\sin ^{2} A}}\\)
\n[\u2235 (a + b)(a + b) = (a + b)2]
\n(a – b)(a + b) = a? — 62)
\n= \\(\\sqrt{\\frac{(1+\\sin A)^{2}}{\\cos ^{2} A}}\\)
\n= \\(\\frac{1+\\sin A}{\\cos A}=\\frac{1}{\\cos A}+\\frac{\\sin A}{\\cos A}\\)
\n= sec A + tan A = R.H.S.<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 4.
\n\\(\\frac{1-\\tan ^{2} A}{\\cot ^{2} A-1}\\) = tan2<\/sup> A
\n\u0c38\u0c3e\u0c27\u0c28.
\nL.H.S. = \\(\\frac{1-\\tan ^{2} A}{\\cot ^{2} A-1}\\)
\n= \\(\\frac{1-\\tan ^{2} \\mathrm{~A}}{\\frac{1}{\\tan ^{2}}-1}\\)
\n= \\(\\frac{1-\\tan ^{2} A}{\\frac{1 \\tan ^{2} A}{\\tan ^{2} A}}\\)
\n= 1 – tan2<\/sup> A \u00d7 \\(\\frac{\\tan ^{2} A}{1-\\tan ^{2} A}\\)
\n= tan2<\/sup> A<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 5.
\n\\(\\frac{1}{\\cos \\theta}\\) – cos \u03b8 = tan \u03b8 . sin \u03b8 \u0c1a\u0c42\u0c2a\u0c02\u0c21\u0c3f
\n\u0c38\u0c3e\u0c27\u0c28.
\nL.H.S = \\(\\frac{1}{\\cos \\theta}\\) – cos \u03b8
\n= \\(\\frac{1-\\cos ^{2} \\theta}{\\cos \\theta}\\)
\n= \\(\\frac{\\sin ^{2} \\theta}{\\cos \\theta}\\) [\u2235 1 – cos2<\/sup> \u03b8 = sin \u03b8]
\n= \\(\\frac{\\sin \\theta \\times \\sin \\theta}{\\cos \\theta}\\)
\n= \\(\\frac{\\sin \\theta}{\\cos \\theta}\\) [\u2235 \\(\\frac{\\sin \\theta}{\\cos \\theta}\\) = tan \u03b8].<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 6.
\nsec A (1 – sin A) (sec A + tan A) \u0c38\u0c42\u0c15\u0c4d\u0c37\u0c4d\u0c2e\u0c40\u0c15\u0c30\u0c3f\u0c02\u0c1a\u0c02\u0c21\u0c3f.
\n\u0c38\u0c3e\u0c27\u0c28.
\nL.H.S. = sec A (1 – sin A) (sec A + tan A)
\n= (sec A – sec A. sin A) . (sec A + tan A)
\n= (sec A – \\(\\frac{1}{\\cos A}\\) . sin A) (sec A + tan A)
\n= (sec A – tan A) (sec A + tan A)
\n= sec2<\/sup> A – tan2<\/sup> A
\n= 1
\n[\u2235 sec2<\/sup> A – tan2<\/sup> A = 1]<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 7.
\n(sin A + cosec A)2<\/sup> + (cos A + sec A)2<\/sup> = 7 + tan2<\/sup> A + cot2<\/sup> A
\n\u0c38\u0c3e\u0c27\u0c28.
\nL.H.S. = (sin A + cosec A)2<\/sup> + (cos A + sec A)2<\/sup>
\n= (sin2<\/sup> A + cosec2<\/sup> A + 2 sin A. cosec A) + (cos2<\/sup> A + sec2<\/sup> A +
\n2 cos A . sec A)
\n[\u2235 (a + b)2<\/sup> = a2<\/sup> + b2<\/sup> + 2ab]
\n= (sin2<\/sup> A + cos2<\/sup> A) + cosec2<\/sup> A + 2 sinA. \\(\\frac{1}{\\sin A}\\) + sec2<\/sup> A + 2 cos A \u00b7 \\(\\frac{1}{\\cos A}\\)
\n[\u2235 \\(\\frac{1}{\\sin A}\\) = cosec A; \\(\\frac{1}{\\cos A}\\) = sec A]
\n= 1 + (1 + cot2<\/sup> A) + 2 + (1 + tan2<\/sup> A) + 2
\n[\u2235 sin2<\/sup> A + cos2<\/sup> A = 1]
\ncosec2<\/sup> A = 1 + cot2<\/sup> A
\nsec2<\/sup> A = 1 + tan2<\/sup> A]
\n= 7 + tan2<\/sup> A + cot2<\/sup> A = R.H.S.<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 8.
\n(1 – cos \u03b8) (1 + cos \u03b8) (1 + cot2<\/sup> \u03b8) \u0c38\u0c42\u0c15\u0c4d\u0c37\u0c4d\u0c2e\u0c40\u0c15\u0c30\u0c3f\u0c02\u0c1a\u0c02\u0c21\u0c3f.
\n\u0c38\u0c3e\u0c27\u0c28.
\n(1 – cos \u03b8) (1 + cos \u03b8) (1 + cot2<\/sup> \u03b8) = (1 – cos2<\/sup> \u03b8) (1 + cot2<\/sup> \u03b8)
\n[\u2235 (a – b)(a + b) = a2<\/sup> – b2<\/sup>]
\n= sin2<\/sup> \u03b8. cosec2<\/sup> \u03b8
\n[: 1 – cos2<\/sup> \u03b8 = sin2<\/sup> \u03b8
\n1 + cot2<\/sup> \u03b8 = cosec2<\/sup> \u03b8]
\n= sin2<\/sup> \u03b8 . \\(\\frac{1}{\\sin ^{2} \\theta}\\) [\u2235 cosec \u03b8 = sin \u03b8]
\n= 1<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 9.
\nsec \u03b8 + tan \u03b8 = p \u0c10\u0c24\u0c47 sec \u03b8 – tan \u03b8 \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24?
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u0c26\u0c24\u0c4d\u0c24\u0c3e\u0c02\u0c36\u0c2e\u0c41
\nsec \u03b8 + tan \u03b8 = p
\nsec2<\/sup> \u03b8 – tan2<\/sup> \u03b8 = 1 \u0c05\u0c28\u0c41 \u0c38\u0c30\u0c4d\u0c35\u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c02 \u0c26\u0c4d\u0c35\u0c3e\u0c30\u0c3e
\nsec2<\/sup> \u03b8 – tan2<\/sup> \u03b8 = (sec \u03b8 + tan \u03b8) (sec \u03b8 – tan \u03b8) = 1
\n= p (sec \u03b8 – tan \u03b8) = 1
\n(\u0c26\u0c24\u0c4d\u0c24\u0c3e\u0c02\u0c36\u0c2e\u0c41 \u0c28\u0c41\u0c02\u0c21\u0c3f)
\n\u21d2 sec \u03b8 – tan \u03b8 = \\(\\frac{1}{p}\\)<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 10.
\ncosec \u03b8 + cot \u03b8 = k \u0c10\u0c24\u0c47 \u0441os \u03b8 = \\(\\frac{k^{2}-1}{k^{2}+1}\\) \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24?
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u0c2a\u0c26\u0c4d\u0c27\u0c24\u0c3f – I:
\n\u0c26\u0c24\u0c4d\u0c24\u0c3e\u0c02\u0c36\u0c2e\u0c41 cosec \u03b8 + cot \u03b8 = k
\nR.H.S. = \\(\\frac{k^{2}-1}{k^{2}+1}\\)<\/p>\n

\"AP<\/p>\n

= \\(\\frac{2 \\cot \\theta(\\cot \\theta+cosec \\theta)}{2 cosec \\theta(\\cot \\theta+cosec \\theta)}\\)<\/p>\n

= \\(\\frac{\\cot \\theta}{cosec \\theta}\\)<\/p>\n

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

= \\(\\frac{\\cos \\theta}{\\sin \\theta} \\times \\frac{\\sin \\theta}{1}\\)
\n= cos \u03b8 = L.H.S.<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c26\u0c4d\u0c27\u0c24\u0c3f – II:
\n\u0c26\u0c24\u0c4d\u0c24\u0c3e\u0c02\u0c36\u0c2e\u0c41 : cosec \u03b8 + cot \u03b8 = k ………………(1)
\n\u0c38\u0c30\u0c4d\u0c35 \u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c02 cosec2<\/sup> \u03b8 – cot2<\/sup> \u03b8 = 1
\n\u21d2 (cosec \u03b8 + cot \u03b8) (cosec \u03b8 – cot \u03b8) = 1
\n[\u2235 a2<\/sup> – b2<\/sup> = (a – b)(a + b)]
\n\u21d2 k (cosec \u03b8 – cot \u03b8) = 1
\n\u21d2 cosec \u03b8 – cot \u03b8 = \\(\\frac{1}{\\mathrm{k}}\\) ……………… (2)
\n(1) \u0c2e\u0c30\u0c3f\u0c2f\u0c41 (2) \u0c32\u0c28\u0c41 \u0c38\u0c3e\u0c27\u0c3f\u0c02\u0c1a\u0c17\u0c3e<\/p>\n

\"AP<\/p>\n","protected":false},"excerpt":{"rendered":"

SCERT AP 10th Class Maths Textbook Solutions Chapter 11 \u0c24\u0c4d\u0c30\u0c3f\u0c15\u0c4b\u0c23\u0c2e\u0c3f\u0c24\u0c3f Exercise 11.4 Textbook Exercise Questions and Answers. AP State Syllabus 10th Class Maths Solutions 11th Lesson \u0c24\u0c4d\u0c30\u0c3f\u0c15\u0c4b\u0c23\u0c2e\u0c3f\u0c24\u0c3f Exercise 11.4 \u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 1. \u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c35\u0c3e\u0c1f\u0c3f\u0c28\u0c3f \u0c38\u0c42\u0c15\u0c4d\u0c37\u0c4d\u0c2e\u0c3f\u0c15\u0c30\u0c3f\u0c02\u0c1a\u0c02\u0c21\u0c3f: (i) (1 + tan \u03b8 + sec \u03b8) (1 + cot \u03b8 – cosec \u03b8) \u0c38\u0c3e\u0c27\u0c28. (1 + tan \u03b8 + … 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":[2],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/21326"}],"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=21326"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/21326\/revisions"}],"predecessor-version":[{"id":21327,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/21326\/revisions\/21327"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/media?parent=21326"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/categories?post=21326"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/tags?post=21326"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}