CBSE Class 12-science Answered

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Asked by Varsneya Srinivas | 15 Mar, 2018, 02:51: PM

Expert Answer

begin mathsize 16px style tan to the power of negative 1 end exponent open parentheses fraction numerator 3 sin 2 straight alpha over denominator 5 plus 3 cos 2 straight alpha end fraction close parentheses plus tan to the power of negative 1 end exponent open parentheses 1 fourth tanα close parentheses space space space space space space space space space space space space space space space space space space space where space minus straight pi over 2 less than straight alpha less than straight pi over 2 Using space sin 2 straight A equals fraction numerator 2 tanA over denominator 1 plus tan squared straight A end fraction space and space cos 2 straight A equals fraction numerator 1 minus tan squared straight A over denominator 1 plus tan squared straight A end fraction rightwards double arrow tan to the power of negative 1 end exponent open parentheses fraction numerator 3 cross times fraction numerator 2 tanα over denominator 1 plus tan squared straight alpha end fraction over denominator 5 plus 3 cross times fraction numerator 1 minus tan squared straight A over denominator 1 plus tan squared straight A end fraction end fraction close parentheses plus tan to the power of negative 1 end exponent open parentheses 1 fourth tanα close parentheses space rightwards double arrow tan to the power of negative 1 end exponent open parentheses fraction numerator fraction numerator 6 tanα over denominator 1 plus tan squared straight alpha end fraction over denominator fraction numerator 5 left parenthesis 1 plus tan squared straight alpha right parenthesis plus 3 open parentheses 1 minus tan squared straight alpha close parentheses over denominator 1 plus tan squared straight alpha end fraction end fraction close parentheses plus tan to the power of negative 1 end exponent open parentheses 1 fourth tanα close parentheses space space rightwards double arrow tan to the power of negative 1 end exponent open parentheses fraction numerator 6 tanα over denominator 5 plus 5 tan squared straight alpha plus 3 minus 3 tan squared straight alpha end fraction close parentheses plus tan to the power of negative 1 end exponent open parentheses 1 fourth tanα close parentheses rightwards double arrow tan to the power of negative 1 end exponent open parentheses fraction numerator 6 tanα over denominator 2 tan squared straight alpha plus 8 end fraction close parentheses plus tan to the power of negative 1 end exponent open parentheses 1 fourth tanα close parentheses Using space tan to the power of negative 1 end exponent plus tan to the power of negative 1 end exponent straight y equals tan to the power of negative 1 end exponent open parentheses fraction numerator straight x plus straight y over denominator 1 minus xy end fraction close parentheses space and space solving space it space further you space get space rightwards double arrow tan to the power of negative 1 end exponent left parenthesis tanα right parenthesis rightwards double arrow straight alpha end style

Answered by Sneha shidid | 16 Mar, 2018, 09:35: AM

CBSE Class 12-science Answered

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CBSE Class 12-science Answered

Concept Videos

Properties of Inverse Trigonometry Part - 1

Properties of Inverse Trigonometry Part - 2

Problem-solving on inverse trigonometric applications ...

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