{"id":3858,"date":"2023-12-12T06:39:12","date_gmt":"2023-12-12T01:09:12","guid":{"rendered":"https:\/\/wordpress-505192-1602719.cloudwaysapps.com\/?p=3858"},"modified":"2023-12-13T17:48:41","modified_gmt":"2023-12-13T12:18:41","slug":"samacheer-kalvi-11th-maths-solutions-chapter-3-ex-3-7","status":"publish","type":"post","link":"https:\/\/samacheerkalviguru.com\/samacheer-kalvi-11th-maths-solutions-chapter-3-ex-3-7\/","title":{"rendered":"Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.7"},"content":{"rendered":"

You can Download Samacheer Kalvi 11th Maths Book Solutions<\/a> Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.<\/p>\n

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.7<\/h3>\n

Question 1.
\nIf A + B + C = 180\u00b0, prove that
\n(i) sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C
\n\"Samacheer
\n(iii) sin2<\/sup> A + sin2<\/sup> B + sin2<\/sup> C = 2 + 2 cos A cos B cos C
\n(iv) sin2<\/sup> A + sin2<\/sup> B – sin2<\/sup> C = 2 sin A sin B cos C
\n\"Samacheer
\n(vii) sin(B + C – A) + sin(C + A – B) + sin(A + B – C) = 4 sin A sin B sin C.
\nSolution:
\n(i)
\nLHS = (sin 2A + sin 2B) + sin 2C
\n= 2 sin (A + B) cos (A – B) + 2 sin C cos C
\n[sin (A + B) = sin (180\u00b0 – C) = sin C]
\n= 2 sin C cos (A – B) + 2 sin C cos C
\n= 2 sin C [ cos (A – B) + cos C]
\n{cos C = cos [180\u00b0 – (A + B)] = – cos (A + B)}
\n= 2 sin C [cos (A – B) – cos (A + B)]
\n\"Samacheer<\/p>\n

(ii)
\n\"Samacheer<\/p>\n

(iii)
\n\"Samacheer
\n[cos (180\u00b0 –\u00a0 C) – cos C – cos C]
\n= 2 + cos C [cos (A – B) + cos (A + B)]
\n= 2+ cos C[2 cos A cos B]
\n= 2 + 2 cos A cos B cos C = RHS<\/p>\n

(iv)
\n\"Samacheer<\/p>\n

(v)
\n\"Samacheer<\/p>\n

(vi)
\n\"Samacheer
\n\"Samacheer<\/p>\n

(vii)
\nNow A + B + C = 180\u00b0
\nSo B + C = 180\u00b0 – A
\nsin (B + C – A) = sin (180\u00b0 – A – A)
\n= sin(180\u00b0 – 2A) = sin 2A
\nNow LHS = sin 2A + sin 2B + sin 2C
\n= 4 sin A sin B sin C (from (i) ans) = RHS<\/p>\n

\"Samacheer<\/p>\n

Question 2.
\nIf A + B + C = 2s, then prove that sin(s – A) sin(s – B) + sin s sin(s – C) = sin A sin B.
\nSolution:
\n\"Samacheer<\/p>\n

Question 3.
\n\"Samacheer
\nSolution:
\n\"Samacheer
\n\u21d2 A+B+C = 180\u00b0
\n\u21d2 A + B = 180\u00b0 – C
\nmultiply 2 on both sides \u21d2 2A + 2B = 360\u00b0 – 2C
\n2(A + B) =360\u00b0 – 2C
\n\u21d2 tan(2A + 2B) = tan(360\u00b0 – 2C) = – tan 2C
\n\"Samacheer
\n\u21d2 tan 2A + tan 2B = -tan2C[1 – tan 2A tan 2B]
\n\u21d2 tan 2A + tan 2B = -tan 2C + tan 2A tan 2B tan 2C
\n\u21d2 tan 2A + tan 2B + tan 2C = tan 2A tan 2B tan 2C
\n\"Samacheer<\/p>\n

Question 4.
\nIf A + B + C = \\(\\frac{\\pi}{2}\\), prove the following
\n(i) sin 2A + sin 2B + sin 2C = 4 cos A cos B cos C
\n(ii) COS 2A + cos 2B + cos 2C = 1 + 4 sin A sin B sin C.
\nSolution:
\n(i) LHS = (sin 2A + sin 2B) + sin 2C
\n= 2 sin (A + B) cos (A – B) + 2 sin C cos C = 2 sin (90\u00b0 – C) cos (A – B) + 2 sin C cos C
\n= 2 cos C [cos (A – B) + sin C] + cos (A + B) ( \u2234 A + B = \u03c0\/2 – C)
\n= 2 cos C [cos (A – B) + cos (A + B)]
\n= 2 cos C [2 cos A cos B]
\n= 4 cos A cos B cos C = RHS<\/p>\n

\"Samacheer<\/p>\n

(ii) LHS = (cos 2A + cos 2B) + cos 2C
\n= 2 cos (A + B) cos (A – B) + 1 – 2 sin2<\/sup> C
\n= 1 + 2 sin C (cos (A – B) – 2 sin2<\/sup> C)
\n{\u2234 cos (A + B) = cos (90\u00b0 – C) = sin C}
\n= 1 + 2 sin C [cos (A- B) – sin C]
\n= 1 + 2 sin C [cos (A – B) – cos (A + B)]
\n= 1 + 2 sin C [2 sin A sin B]
\n= 1 + 4 sin A sin B sin C = RHS<\/p>\n

Question 5.
\nIf \u2206ABC is a right triangle and if \u2220A = \\(\\frac{\\pi}{2}\\), then prove that
\n(i) cos2 <\/sup>B + cos2 <\/sup>C = 1
\n(ii) sin2 <\/sup>B + sin2 <\/sup>C = 1
\n\"Samacheer
\nSolution:
\n\"Samacheer
\n\"Samacheer
\n\"Samacheer<\/p>\n

Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.7 Additional Questions<\/h3>\n

Question 1.
\nA + B + C = \u03c0, prove that sin 2A – sin 2B + sin 2C = 4 cos A sin B cos C
\nSolution:
\nLHS = sin 2A – sin 2B + sin 2C
\n= sin 2A + sin 2C – sin 2B
\n\"Samacheer
\n= 2 sin (A + C) cos (A – C) – 2 sin B cos B
\n= 2 sin (180\u00b0 – B) cos (A-C) – 2 sin B cos B
\n= 2 sin B cos (A – C) – 2 sin B cos B
\n= 2 sin B [cos (A – C) – cos B]
\n= 2 sin B [cos (A – C) – cos (180\u00b0 – (A + C))]
\n= 2 sin B [cos (A – C) + cos (A + C)]
\n= 2 sin B [2 cos A cos C]
\n= 4 cos A sin B cos C = RHS<\/p>\n

\"Samacheer<\/p>\n

Question 2.
\n\"Samacheer
\nSolution:
\n\"Samacheer
\n\"Samacheer
\nsubstitute in (1) we get,
\n\"Samacheer<\/p>\n","protected":false},"excerpt":{"rendered":"

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations. Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.7 Question 1. If A + B + C = 180\u00b0, prove that (i) sin 2A + …<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":"","footnotes":""},"categories":[7],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/3858"}],"collection":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/comments?post=3858"}],"version-history":[{"count":1,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/3858\/revisions"}],"predecessor-version":[{"id":60125,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/3858\/revisions\/60125"}],"wp:attachment":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/media?parent=3858"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/categories?post=3858"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/tags?post=3858"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}