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Let 0^(@)ltthetalt45^(@).Which one of the following is correct ? |
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Answer» `sin^(2)theta+cos^(6)theta=sin^(6)theta+cos^(2)theta` `sin^(2)theta+cos^(6)theta=sin^(6)theta+cos^(2)theta` `rArrsin^(6)theta-cos^(6)theta=(sin^(2)theta)^(2)-(cos^(2)theta)^(2)` Taking L . H . S, `=(sin^(2)theta-cos^(2)theta)(sin^(4)theta+sin^(2)thetacos^(2)theta+cos^(4)theta)` `(because a^(3)-b^(3)=(a-b)(a^(2)+ab+b^(2)))` `=(sin^(2)theta-cos^(2)theta)(sin^(4)theta+cos^(4)theta+sin^(2)thetacos^(2)theta)` `=(sin^(2)theta-cos^(2)theta)(sin^(2)theta+cos^(2)theta)-2sin^(2)thetacos^(2)theta+sin^(2)thetacos^(2)theta` `=sin^(2)theta-cos^(2)theta-sin^(2)thetacos^(2)theta` Which is not EQUAL to R .H . S.,`sin^(2)theta-cos^(2)theta` Option (a )is not CORRECT . ( b )`therefore"cosec"^(6)theta-cot^(6)theta` `("cosec"^(2)theta-cot^(2)theta)[("cosec"^(2)theta-cot^(2)theta)^(2)+("cosec "thetacottheta)]` `therefore`Option (b ) is also not correct. (c ) `sin^(4)theta+cos^(4)theta=(sin^(2)theta+cos^(2)theta)^(2)-2sin^(2)thetacos^(2)theta` `=1-2sin^(2)thetacos^(2)theta`. Which is not equal to `sin^(2)theta-cos^(2)theta`. Hence , option ( c) is also not correct . (d)`("cosec"^(2)theta+cot^(4)theta)` `="cosec"^(2)theta+("cosec"^(2)theta-1)^(2)` `="cosec"^(2)theta+"cosec"^(4)theta+1-2" cosec"^(2)theta` `="cosec"^(4)theta+1-"cosec"^(2)theta` `="cosec"^(4)theta+cot^(2)theta` Thus option (d ) is correct . |
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