{"id":1063,"date":"2023-11-01T07:05:30","date_gmt":"2023-11-01T01:35:30","guid":{"rendered":"https:\/\/samacheer-kalvi.com\/?p=1063"},"modified":"2023-11-10T11:25:23","modified_gmt":"2023-11-10T05:55:23","slug":"samacheer-kalvi-10th-maths-guide-chapter-6-ex-6-1","status":"publish","type":"post","link":"https:\/\/samacheer-kalvi.com\/samacheer-kalvi-10th-maths-guide-chapter-6-ex-6-1\/","title":{"rendered":"Samacheer Kalvi 10th Maths Guide Chapter 6 Trigonometry Ex 6.1"},"content":{"rendered":"

Students can download Maths Chapter 6 Trigonometry Ex 6.1 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide<\/a> Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.<\/p>\n

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 6 Trigonometry Ex 6.1<\/h2>\n

Question 1.
\nProve the following identities.
\n(i) cot \u03b8 + tan \u03b8 = sec \u03b8 cosec \u03b8
\n(ii) tan4<\/sup> \u03b8 + tan2<\/sup> \u03b8 = sec4<\/sup> \u03b8 – sec2<\/sup> \u03b8
\nAnswer:
\n(i) L. H. S = cot \u03b8 + tan \u03b8
\n= \\(\\frac{\\cos \\theta}{\\sin \\theta}+\\frac{\\sin \\theta}{\\cos \\theta}\\)
\n= \\(\\frac{\\cos ^{2} \\theta+\\sin ^{2} \\theta}{\\sin \\theta \\cos \\theta}\\)
\n[cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1]
\n= \\(\\frac{1}{\\sin \\theta \\cos \\theta}\\)
\n= sec \u03b8 . cosec \u03b8 = R. H. S
\n\u2234 cot \u03b8 + tan \u03b8 = sec \u03b8 cosec \u03b8<\/p>\n

\"Samacheer<\/p>\n

(ii) tan4<\/sup> \u03b8 + tan2<\/sup> \u03b8 = sec4<\/sup> \u03b8 – sec2<\/sup> \u03b8
\nL.H.S = tan4<\/sup> \u03b8 + tan2<\/sup> \u03b8
\n= tan2<\/sup> \u03b8 (tan2<\/sup> \u03b8 + 1)
\n= tan2<\/sup> \u03b8 sec2<\/sup> \u03b8
\nR.H.S = sec4<\/sup> \u03b8 – sec2<\/sup> \u03b8
\n= sec2<\/sup> \u03b8 (sec2<\/sup> \u03b8 – 1)
\n= sec2<\/sup> \u03b8 tan2<\/sup> \u03b8
\nL.H.S = R.H.S
\n\u2234 tan4<\/sup> \u03b8 + tan2<\/sup> \u03b8 = sec4<\/sup> \u03b8 – sec2<\/sup> \u03b8<\/p>\n

Question 2.
\nProve the following identities.
\n(i) \\(\\frac{1-\\tan ^{2} \\theta}{\\cot ^{2} \\theta-1}\\) = tan2<\/sup> \u03b8
\n(ii) \\(\\frac{\\cos \\theta}{1+\\sin \\theta}\\) = sec \u03b8 – tan \u03b8
\nAnswer:
\n(i) \\(\\frac{1-\\tan ^{2} \\theta}{\\cot ^{2} \\theta-1}\\) = tan2<\/sup> \u03b8
\n\"Samacheer
\n(ii) \\(\\frac{\\cos \\theta}{1+\\sin \\theta}\\) = sec \u03b8 – tan \u03b8
\n\"Samacheer
\nAliter:
\nL.H.S = \\(\\frac{\\cos \\theta}{1-\\sin \\theta}\\)
\n[conjugate (1 – sin \u03b8)]
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 3.
\nProve the following identities.
\n\"Samacheer
\nSolution:
\n\"Samacheer
\n\"Samacheer
\n\"Samacheer
\n\"Samacheer<\/p>\n

Question 4.
\nProve the following identities.
\n(i) sec6<\/sup> \u03b8 = tan6<\/sup> \u03b8 + 3 tan2<\/sup> \u03b8 sec2<\/sup>\u00a0\u03b8 + 1
\n(ii) (sin \u03b8 + sec \u03b8)2<\/sup> + (cos \u03b8 + cosec \u03b8)2<\/sup> = 1 + (sec \u03b8 + cosec \u03b8)2<\/sup>
\nAnswer:
\n(i) sec6<\/sup> \u03b8 = tan6<\/sup> \u03b8 + 3 tan2<\/sup> \u03b8 sec2<\/sup> \u03b8 + 1
\nL.H.S = sec6<\/sup> \u03b8
\n= (sec2<\/sup> \u03b8)3<\/sup> = (1 + tan2<\/sup> \u03b8)3<\/sup>
\n= 1 + (tan2<\/sup> \u03b8)3<\/sup> + 3 (1) (tan2<\/sup> \u03b8) (1 + tan2<\/sup> \u03b8) [(a + b)3<\/sup> = a3<\/sup> + b3<\/sup> + 3 ab (a + b)]
\n= 1 + tan6<\/sup> \u03b8 + 3 tan2<\/sup> \u03b8(1 + tan2<\/sup> \u03b8)
\n= 1 + tan6<\/sup> \u03b8 + 3 tan2<\/sup> \u03b8 (sec2<\/sup> \u03b8)
\n= 1 + tan6<\/sup> \u03b8 + 3 tan2<\/sup> \u03b8 sec2<\/sup> \u03b8
\n= tan6<\/sup> \u03b8 + 3 tan2<\/sup> \u03b8 sec2<\/sup> \u03b8 + 1
\nL.H.S = R.H.S<\/p>\n

(ii) (sin \u03b8 + sec \u03b8)2<\/sup> + (cos \u03b8 + cosec \u03b8)2<\/sup> = 1 + (sec \u03b8 + cosec \u03b8)2<\/sup>
\nL.H.S = (sin \u03b8 + sec \u03b8)2<\/sup> + (cos \u03b8 + cosec \u03b8)2<\/sup>]
\n= [sin2<\/sup> \u03b8 + sec2<\/sup> \u03b8 + 2 sin \u03b8 sec \u03b8 + cos2<\/sup> \u03b8 + cosec2<\/sup> \u03b8 + 2 cos \u03b8 cosec \u03b8]
\n= (sin2<\/sup> \u03b8 + cos2<\/sup> \u03b8) + (sec2<\/sup> \u03b8 + cosec2<\/sup> \u03b8) + 2 (sin \u03b8 sec \u03b8 + cos \u03b8 cosec \u03b8)
\n\"Samacheer
\n= 1 + sec2<\/sup> \u03b8 + cosec2<\/sup> \u03b8 + 2 sec \u03b8 cosec \u03b8
\n= 1 + (sec \u03b8 + cosec \u03b8)2<\/sup>
\nL.H.S = R.H.S
\n\u2234 (sin \u03b8 + sec \u03b8)2<\/sup> + (cos \u03b8 + cosec \u03b8)2<\/sup> = 1 + (sec \u03b8 + cosec \u03b8)2<\/sup><\/p>\n

\"Samacheer<\/p>\n

Question 5.
\nProve the following identities.
\n(i) sec4<\/sup> \u03b8 (1 – sin4<\/sup> \u03b8) – 2 tan2<\/sup>\u00a0\u03b8 = 1
\n(ii) \\(\\frac{\\cot \\theta-\\cos \\theta}{\\cot \\theta+\\cos \\theta}=\\frac{\\csc \\theta-1}{\\csc \\theta+1}\\)
\nAnswer:
\n(i) L.H.S = sec4<\/sup> \u03b8 (1 – sin4<\/sup> \u03b8) – 2 tan2<\/sup> \u03b8
\n\"Samacheer
\nL.H.S = R.H.S
\n\u2234 sec4<\/sup> \u03b8 (1 – sin4<\/sup> \u03b8) – 2 tan2<\/sup> \u03b8 = 1<\/p>\n

(ii) \\(\\frac{\\cot \\theta-\\cos \\theta}{\\cot \\theta+\\cos \\theta}=\\frac{\\csc \\theta-1}{\\csc \\theta+1}\\)
\n\"Samacheer
\n\"Samacheer
\n\"Samacheer<\/p>\n

Question 6.
\nProve the following identities.
\n\"Samacheer
\n\"Samacheer
\nAnswer:
\n\"Samacheer
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 7.
\n(i) If sin \u03b8 + cos \u03b8 = \\(\\sqrt { 3 }\\), then prove that tan \u03b8 + cot \u03b8 = 1.
\n(ii) If \\(\\sqrt { 3 }\\) sin \u03b8 – cos \u03b8 = \u03b8, then show that tan 3\u03b8 = \\(\\frac{3 \\tan \\theta-\\tan ^{3} \\theta}{1-3 \\tan ^{2} \\theta}\\)
\nAnswer:
\nsin \u03b8 + cos \u03b8 = \\(\\sqrt { 3 }\\) (squaring on both sides)
\n(sin \u03b8 + cos \u03b8)2<\/sup> = (\\(\\sqrt { 3 }\\))2<\/sup>
\nsin2<\/sup> \u03b8 + cos2<\/sup> \u03b8 + 2 sin \u03b8 cos \u03b8 = 3
\n1 + 2 sin \u03b8 cos \u03b8 = 3
\n2 sin \u03b8 cos \u03b8 = 3 – 1
\n2 sin \u03b8 cos \u03b8 = 2
\n\u2234 sin \u03b8 cos \u03b8 = 1
\nL.H.S = tan \u03b8 + cot \u03b8
\n\"Samacheer
\nL.H.S = R.H.S \u21d2 tan \u03b8 + cot \u03b8 = 1<\/p>\n

(ii) If \\(\\sqrt { 3 }\\) sin \u03b8 – cos \u03b8 = 0
\nTo prove tan 3\u03b8 = \\(\\frac{3 \\tan \\theta-\\tan ^{3} \\theta}{1-3 \\tan ^{2} \\theta}\\)
\n\\(\\sqrt { 3 }\\) sin \u03b8 – cos \u03b8 = 0
\n\\(\\sqrt { 3 }\\) sin \u03b8 = cos \u03b8
\n\\(\\frac{\\sin \\theta}{\\cos \\theta}=\\frac{1}{\\sqrt{3}}\\)
\ntan \u03b8 = tan 30\u00b0
\n\u03b8 = 30\u00b0
\nL.H.S = tan 3\u03b8\u00b0
\n= tan3 (30\u00b0)
\n= tan 90\u00b0
\n= undefined (\u221d)
\n\"Samacheer
\n\u2234 tan 3\u03b8 = \\(\\frac{3 \\tan \\theta-\\tan ^{3} \\theta}{1-3 \\tan ^{2} \\theta}\\)<\/p>\n

\"Samacheer<\/p>\n

Question 8.
\n(i) If \\(\\frac{\\cos \\alpha}{\\cos \\beta}=m\\) and \\(\\frac{\\cos \\alpha}{\\cos \\beta}=n\\) then prove that (m2 + n2) cos2 \u03b2 = n2
\n(ii) If cot \u03b8 + tan \u03b8 = x and sec \u03b8 – sec \u03b8 – cos \u03b8 = y, then prove that (x2y)2\/3 – (xy2)2\/3 = 1
\nAnswer:
\n(i) L.H.S = (m2<\/sup> + n2<\/sup>) cos2<\/sup> \u03b2
\n\"Samacheer
\nL.H.S = R.H.S \u21d2 \u2234 (m2<\/sup> + n2<\/sup>) cos2<\/sup> \u03b2 = n2<\/sup><\/p>\n

(ii) Given cot \u03b8 + tan \u03b8 = x sec \u03b8 – cos \u03b8 = y
\nx = cot \u03b8 + tan \u03b8
\nx = \\(\\frac{1}{\\tan \\theta}\\) + tan \u03b8
\n= \\(\\frac{1+\\tan ^{2} \\theta}{\\tan \\theta}\\) = \\(\\frac{\\sec ^{2} \\theta}{\\tan \\theta}\\)
\n\"Samacheer
\ny = sec \u03b8 – cos \u03b8
\n= \\(\\frac{1}{\\cos \\theta}-\\cos \\theta=\\frac{1-\\cos ^{2} \\theta}{\\cos \\theta}\\)
\ny = \\(\\frac{\\sin ^{2} \\theta}{\\cos \\theta}\\)
\n\"Samacheer<\/p>\n

Question 9.
\n(i) If sin \u03b8 + cos \u03b8 = p and sec \u03b8 + cosec \u03b8 = q, then prove that q (p2<\/sup> – 1) = 2 p
\n(ii) If sin \u03b8 (1 + sin2<\/sup> \u03b8) = cos2<\/sup> \u03b8, then prove that cos6<\/sup> \u03b8 – 4 cos4<\/sup> \u03b8 + 8 cos2<\/sup> \u03b8 = 4
\nAnswer:
\n(i) p = sin \u03b8 + cos \u03b8
\np2<\/sup> = (sin \u03b8 + cos \u03b8)2<\/sup>
\n= sin2<\/sup> \u03b8 + cos2<\/sup> \u03b8 + 2 sin \u03b8 cos \u03b8 = 1 + 2 sin \u03b8 cos \u03b8
\nq = sec \u03b8 + cosec \u03b8
\n= \\(\\frac{1}{\\cos \\theta}+\\frac{1}{\\sin \\theta}=\\frac{\\sin \\theta+\\cos \\theta}{\\sin \\theta \\cos \\theta}\\)
\nL.H.S = q(p2<\/sup> – 1)
\n\"Samacheer<\/p>\n

(ii) sin \u03b8 (1 + sin2<\/sup> \u03b8) = cos2<\/sup> \u03b8
\nsin \u03b8 (1 + 1 – cos2<\/sup> \u03b8) = cos2<\/sup> \u03b8
\nsin \u03b8 (2 – cos2<\/sup> \u03b8) = cos2<\/sup> \u03b8
\nSquaring on both sides,
\nsin2<\/sup> \u03b8 (2 – cos2<\/sup> \u03b8)2<\/sup> = cos4<\/sup> \u03b8
\n(1 – cos2<\/sup> \u03b8) (4 + cos4<\/sup> \u03b8 – 4 cos2<\/sup> \u03b8) = cos4<\/sup> \u03b8
\n4 cos4<\/sup> \u03b8 – 4 cos2<\/sup> \u03b8 – cos6<\/sup> \u03b8 + 4 cos4<\/sup> \u03b8 = cos4<\/sup> \u03b8
\n4 + 5 cos4<\/sup> \u03b8 – 8 cos2<\/sup> \u03b8 – cos6<\/sup> \u03b8 = cos4<\/sup> \u03b8
\n– cos6<\/sup> \u03b8 + 5 cos4<\/sup> \u03b8 – cos4<\/sup> \u03b8 – 8 cos2<\/sup> \u03b8 = -4
\n– cos6<\/sup> \u03b8 + 4 cos4<\/sup> \u03b8 – 8 cos2<\/sup> \u03b8 = -4
\ncos6<\/sup> \u03b8 – 4 cos4<\/sup> \u03b8 + 8 cos2<\/sup> \u03b8 = 4
\nHence it is proved<\/p>\n

\"Samacheer<\/p>\n

Question 10.
\nIf \\(\\frac{\\cos \\theta}{1+\\sin \\theta}\\) = \\(\\frac { 1 }{ a } \\), then prove that \\(\\frac{a^{2}-1}{a^{2}+1}\\) = sin \u03b8
\nAnswer:
\n\"Samacheer
\n\"Samacheer<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can download Maths Chapter 6 Trigonometry Ex 6.1 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams. Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 6 Trigonometry Ex 6.1 Question …<\/p>\n","protected":false},"author":3,"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:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/posts\/1063"}],"collection":[{"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/comments?post=1063"}],"version-history":[{"count":1,"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/posts\/1063\/revisions"}],"predecessor-version":[{"id":49365,"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/posts\/1063\/revisions\/49365"}],"wp:attachment":[{"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/media?parent=1063"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/categories?post=1063"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/tags?post=1063"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}