2<\/sup> \u03b8 = 1<\/p>\n(ii) \\(\\frac{\\cot \\theta-\\cos \\theta}{\\cot \\theta+\\cos \\theta}=\\frac{\\csc \\theta-1}{\\csc \\theta+1}\\)
\n
\n
\n<\/p>\n
Question 6.
\nProve the following identities.
\n
\n
\nAnswer:
\n
\n<\/p>\n
<\/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
\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
\n\u2234 tan 3\u03b8 = \\(\\frac{3 \\tan \\theta-\\tan ^{3} \\theta}{1-3 \\tan ^{2} \\theta}\\)<\/p>\n
<\/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
\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
\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<\/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<\/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<\/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
\n<\/p>\n","protected":false},"excerpt":{"rendered":"
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