{"id":11981,"date":"2022-09-08T23:07:14","date_gmt":"2022-09-08T17:37:14","guid":{"rendered":"https:\/\/mathemerize.com\/?p=11981"},"modified":"2022-09-08T23:07:19","modified_gmt":"2022-09-08T17:37:19","slug":"prove-the-following-identities-where-the-angles-involved-are-acute-angles-for-which-the-expressions-are-defined","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/prove-the-following-identities-where-the-angles-involved-are-acute-angles-for-which-the-expressions-are-defined\/","title":{"rendered":"Prove the following identities, where the angles involved are acute angles for which the expressions are defined :"},"content":{"rendered":"

Question :<\/h2>\n

(i)<\/strong>\u00a0 \\(1 – cos\\theta\\over 1 + cos\\theta\\) = \\(({cosec\\theta – cot\\theta})^2\\)<\/p>\n

(ii)<\/strong>\u00a0 \\(1 + sin\\theta\\over cos\\theta\\) + \\(cos\\theta\\over 1 + sin\\theta\\) = \\(2sec\\theta\\)<\/p>\n

(iii)<\/strong>\u00a0 \\(tan\\theta\\over 1 – cot\\theta\\) + \\(cot\\theta\\over 1 – tan\\theta\\) = \\(sec\\theta + cosec\\theta\\) + 1<\/p>\n

(iv)<\/strong>\u00a0 \\(1 + sec\\theta\\over sec\\theta\\) = \\(sin^2\\theta\\over 1 – cos\\theta\\)<\/p>\n

(v)<\/strong>\u00a0 \\(cos A – sin A + 1\\over cos A + sin A – 1\\) = cosec A + cot A<\/p>\n

(vi)<\/strong>\u00a0 \\(\\sqrt{1 + sin\\alpha\\over 1 – sin\\alpha}\\) = \\(sec\\alpha + tan\\alpha\\)<\/p>\n

(vii)<\/strong>\u00a0 \\(sin\\theta – 2 sin^3\\theta\\over 2cos^3\\theta – cos\\theta\\) = \\(tan\\theta\\)<\/p>\n

(viii)<\/strong>\u00a0 \\(({sin A + cosec A})^2\\) + \\(({cos A + sec A})^2\\) = 7 + \\(tan^2 A\\) + \\(cot^2A\\)<\/p>\n

(ix)<\/strong>\u00a0 (cosec A – sin A)(sec A – cos A) = \\(1\\over tan A + cot A\\)<\/p>\n

(x)<\/strong>\u00a0 \\(1 + tan^2\\theta\\over 1 + cot^2\\theta\\) = \\(({1 – tan\\theta\\over 1 – cot\\theta})^2\\)<\/p>\n

Solution :<\/h2>\n

(i)\u00a0 R.H.S<\/strong> = \\(({cosec\\theta – cot\\theta})^2\\) = \\(({1\\over sin\\theta} – {cos\\theta\\over sin\\theta})^2\\)<\/p>\n

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

= \\(1 – cos\\theta\\over 1 + cos\\theta\\) = L.H.S\u00a0<\/strong><\/p>\n

(ii)\u00a0 L.H.S<\/strong> = \\(1 + sin\\theta\\over cos\\theta\\) + \\(cos\\theta\\over 1 + sin\\theta\\) = \\(({1 + sin\\theta})^2 + cos^2\\theta\\over cos\\theta(1 + sin\\theta)\\)<\/p>\n

= \\(1 + sin^2\\theta + 2sin\\theta + cos^2\\theta\\over cos\\theta(1 + sin\\theta)\\)<\/p>\n

= \\(1 + 1 + 2sin\\theta\\over cos\\theta(1 + sin\\theta)\\) = \\(2(1 + sin\\theta)\\over cos\\theta(1 + sin\\theta)\\) = \\(2\\over cos\\theta\\)<\/p>\n

= \\(2\\sec\\theta\\) = R.H.S.<\/strong><\/p>\n

(iii)\u00a0 L.H.S<\/strong> = \\(tan\\theta\\over 1 – cot\\theta\\) + \\(cot\\theta\\over 1 – tan\\theta\\) = \\(tan\\theta\\over 1 – {1\\over tan\\theta}\\) + \\({1\\over tan\\theta}\\over 1 – tan\\theta\\)<\/p>\n

= \\(tan^2\\theta\\over tan\\theta – 1\\) – \\(1\\over tan\\theta(tan\\theta – 1)\\) = \\(tan^3\\theta – 1\\over tan\\theta(tan\\theta – 1)\\)<\/p>\n

= \\({(tan\\theta – 1)(tan^2\\theta + tan\\theta + 1)}\\over tan\\theta(tan\\theta – 1)\\) = \\(sec^2\\theta + tan^2\\theta\\over tan\\theta\\)<\/p>\n

= \\(sec^2\\theta\\over tan\\theta\\) + \\(tan\\theta\\over tan\\theta\\) = \\(sec\\theta cosec\\theta\\) + 1 = R.H.S<\/strong><\/p>\n

(iv)\u00a0 L.H.S<\/strong> = \\(1 + sec\\theta\\over sec\\theta\\) = \\(cos\\theta + 1\\over cos\\theta\\) \\(\\times\\) \\(cos\\theta\\over 1\\)<\/p>\n

= \\(cos\\theta + 1\\) = \\((1 + cos\\theta)(1 – cos\\theta)\\over 1 – cos\\theta\\)<\/p>\n

= \\(1 – cos^2\\theta\\over 1 – cos\\theta\\) = \\(sin^2\\theta\\over 1 – cos\\theta\\) = R.H.S.<\/strong><\/p>\n

(v)\u00a0 L.H.S<\/strong> = \\(cos A – sin A + 1\\over cos A + sin A – 1\\)<\/p>\n

Dividing numerator and denominator by sin A, we get<\/p>\n

L.H.S = \\(cot A + cosec A – 1\\over 1 – cosec A + cot A\\) = \\(cot A + cosec A – (cosec^2 A – cot^2A)\\over 1 – cosec A + cot A\\)<\/p>\n

= \\(cot A + cosec A – (cosec A + cot A)(cosec A – cot A)\\over 1 – cosec A + cot A\\)<\/p>\n

= \\((cot A + cosec A)(1 – cosec A + cot A)\\over 1 – cosec A + cot A\\) = cot A + cosec A = R.H.S<\/strong><\/p>\n

(vi)\u00a0 L.H.S<\/strong> = \\(\\sqrt{1 + sin\\alpha\\over 1 – sin\\alpha}\\) = \\(\\sqrt{(1 + sin\\alpha)(1 + sin\\alpha)\\over (1 – sin\\alpha)(1 + sin\\alpha)}\\)<\/p>\n

= \\(\\sqrt{(1 + sin\\alpha)^2\\over 1 – sin^2\\alpha}\\) = \\(\\sqrt{(1 + sin\\alpha)^2\\over cos^2\\alpha}\\)<\/p>\n

= \\(1 + sin\\alpha\\over cos\\alpha\\) = \\(1\\over cos\\alpha\\) + \\(sin\\alpha\\over cos\\alpha\\) = \\(sec\\alpha + tan\\alpha\\) = R.H.S<\/strong><\/p>\n

(vii)\u00a0 L.H.S<\/strong> = \\(sin\\theta – 2 sin^3\\theta\\over 2cos^3\\theta – cos\\theta\\) = \\(sin\\theta(1 – 2sin^2\\theta)\\over cos\\theta(2cos^2\\theta –\u00a0 1)\\)<\/p>\n

= \\(sin\\theta[1 – 2(1 – cos^2\\theta)]\\over cos\\theta[2cos^2\\theta – 1]\\) = \\(sin\\theta[1 – 2 + 2cos^2\\theta]\\over cos\\theta[2cos^2\\theta – 1]\\)<\/p>\n

= \\(sin\\theta[2cos^2\\theta – 1]\\over cos\\theta[2cos^2\\theta – 1]\\) = \\(sin\\theta\\over cos\\theta\\) = \\(tan\\theta\\) = R.H.S<\/strong>.<\/p>\n

(viii)\u00a0 L.H.S<\/strong> = \\(({sin A + cosec A})^2\\) + \\(({cos A + sec A})^2\\)<\/p>\n

= \\(sin^2A\\) + \\(cosec^2A\\) + 2sin A cosec A + \\(cos^2A\\) + \\(sec^2A\\) + 2cos A.sec A<\/p>\n

= \\(sin^2A\\) + \\(cosec^2A\\) + \\(2sin A\\over sin A\\)\u00a0 + \\(cos^2A\\) + \\(sec^2A\\) + \\(2cos A\\over cos A\\)<\/p>\n

= \\(sin^2A\\) + \\(cosec^2A\\)\u00a0 + \\(cos^2A\\) + \\(sec^2A\\) + 4<\/p>\n

= \\(sin^2A\\) + \\(cos^2A\\) + \\(cosec^2A\\) + \\(sec^2A\\) + 4<\/p>\n

= 1 + (1 +\\(cot^2A\\)) + (1 + \\(tan^2A\\)) + 4<\/p>\n

= 7 + \\(cot^2A\\) + \\(tan^2A\\) = R.H.S<\/strong><\/p>\n

(ix)\u00a0 L.H.S<\/strong> = (cosec A – sin A)(sec A – cos A)<\/p>\n

= cosec A sec A – cosec A cos A – sin A sec A + sin A cos A<\/p>\n

= \\(1\\over sin A cos A\\) – \\(cos A\\over sin A\\) – \\(sin A\\over cos A\\) + sin A cos A<\/p>\n

= \\(1 – cos^2A – sin^2A + sin^2A.cos^2A\\over sin A.cos A\\)<\/p>\n

= \\(1 – (cos^2A + sin^2A) + sin^2A.cos^2A\\over sin A.cos A\\) = = \\(1 – 1 + sin^2A.cos^2A\\over sin A.cos A\\)<\/p>\n

= = \\(sin^2A.cos^2A\\over sin A.cos A\\) = sin A cos A\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0<\/strong>………..(1)<\/p>\n

R.H.S =<\/strong> \\(1\\over tan A + cot A\\) = \\(1\\over {sin A\\over cos A} + {cos A\\over sin A}\\)<\/p>\n

= \\(sin A cos A\\over sin^2A + cos^2A\\) = sin A cos A\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0<\/strong>…………(2)<\/p>\n

From (1) and (2), we have<\/p>\n

L.H.S = R.H.S<\/strong><\/p>\n

(x)\u00a0 L.H.S<\/strong> = \\(1 + tan^2\\theta\\over 1 + cot^2\\theta\\) = \\(sec^2\\theta\\over cosec^2\\theta\\)<\/p>\n

= \\(sin^2\\theta\\over cos^2\\theta\\) = \\(tan^2\\theta\\)<\/strong>\u00a0 \u00a0 \u00a0 ……….(1)<\/p>\n

R.H.S<\/strong> = \\(({1 – tan\\theta\\over 1 – cot\\theta})^2\\) = \\(({1 – tan\\theta\\over 1 – {1\\over tan\\theta}})^2\\)<\/p>\n

= \\(({1 – tan\\theta\\over {tan\\theta – 1\\over tan\\theta}})^2\\) = \\(({1 – tan\\theta\\over -(1 – tan\\theta)}.tan\\theta)^2\\) = \\(tan^2\\theta\\)<\/strong>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0……(2)<\/p>\n

From (1) and (2), clearly, L.H.S = R.H.S<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":"

Question : (i)\u00a0 \\(1 – cos\\theta\\over 1 + cos\\theta\\) = \\(({cosec\\theta – cot\\theta})^2\\) (ii)\u00a0 \\(1 + sin\\theta\\over cos\\theta\\) + \\(cos\\theta\\over 1 + sin\\theta\\) = \\(2sec\\theta\\) (iii)\u00a0 \\(tan\\theta\\over 1 – cot\\theta\\) + \\(cot\\theta\\over 1 – tan\\theta\\) = \\(sec\\theta + cosec\\theta\\) + 1 (iv)\u00a0 \\(1 + sec\\theta\\over sec\\theta\\) = \\(sin^2\\theta\\over 1 – cos\\theta\\) (v)\u00a0 \\(cos A – …<\/p>\n

Prove the following identities, where the angles involved are acute angles for which the expressions are defined :<\/span> Read More »<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-global-header-display":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":""},"categories":[43,60],"tags":[],"yoast_head":"\nProve the following identities, where the angles involved are acute angles for which the expressions are defined : - Mathemerize<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mathemerize.com\/prove-the-following-identities-where-the-angles-involved-are-acute-angles-for-which-the-expressions-are-defined\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Prove the following identities, where the angles involved are acute angles for which the expressions are defined : - 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