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Prove the following identities: \[.\text{ }\left[ \mathbf{1}\text{ }\text{ }\mathbf{cos}\text{ }\mathbf{2x}\text{ }+\text{ }\mathbf{sin}\text{ }\mathbf{2x} \right]\text{ }/\text{ }\left[ \mathbf{1}\text{ }+\text{ }\mathbf{cos}\text{ }\mathbf{2x}\text{ }+\text{ }\mathbf{sin}\text{ }\mathbf{2x} \right]\text{ }=\text{ }\mathbf{tan}\text{ }\mathbf{x}\]
Prove the following identities: \[.\text{ }\left[ \mathbf{1}\text{ }\text{ }\mathbf{cos}\text{ }\mathbf{2x}\text{ }+\text{ }\mathbf{sin}\text{ }\mathbf{2x} \right]\text{ }/\text{ }\left[ \mathbf{1}\text{ }+\text{ }\mathbf{cos}\text{ }\mathbf{2x}\text{ }+\text{ }\mathbf{sin}\text{ }\mathbf{2x} \right]\text{ }=\text{ }\mathbf{tan}\text{ }\mathbf{x}\]
CBSE | Class 11 | Exercise 9.1 | Maths | RD Sharma | Trigonometric Functions at Multiples and Submultiples of an Angle
Prove the following identities: \[.\text{ }\left[ \mathbf{1}\text{ }\text{ }\mathbf{cos}\text{ }\mathbf{2x}\text{ }+\text{ }\mathbf{sin}\text{ }\mathbf{2x} \right]\text{ }/\text{ }\left[ \mathbf{1}\text{ }+\text{ }\mathbf{cos}\text{ }\mathbf{2x}\text{ }+\text{ }\mathbf{sin}\text{ }\mathbf{2x} \right]\text{ }=\text{ }\mathbf{tan}\text{ }\mathbf{x}\]

Take LHS:

\[\left[ 1\text{ }\text{ }cos\text{ }2x\text{ }+\text{ }sin\text{ }2x \right]\text{ }/\text{ }\left[ 1\text{ }+\text{ }cos\text{ }2x\text{ }+\text{ }sin\text{ }2x \right]\]

From formulae, \[cos\text{ }2x\text{ }=\text{ }1\text{ }\text{ }2\text{ }si{{n}^{2}}~x\]

\[=\text{ }2\text{ }co{{s}^{2}}~x\text{ }\text{ }1\]

\[Sin\text{ }2x\text{ }=\text{ }2\text{ }sin\text{ }x\text{ }cos\text{ }x\]

SO,

i

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