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

Take LHS:

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

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

So,

\[sin\text{ }2x\text{ }/\text{ }\left( 1\text{ }\text{ }cos\text{ }2x \right)\text{ }=\text{ }\left( 2\text{ }sin\text{ }x\text{ }cos\text{ }x \right)\text{ }/\text{ }(1\text{ }\text{ }(1\text{ }\text{ }2si{{n}^{2}}~x))\]

\[=\text{ }\left( 2\text{ }sin\text{ }x\text{ }cos\text{ }x \right)\text{ }/\text{ }(1\text{ }\text{ }1\text{ }+\text{ }2si{{n}^{2}}~x)]\]

\[=\text{ }[2\text{ }sin\text{ }x\text{ }cos\text{ }x\text{ }/\text{ }2\text{ }si{{n}^{2}}~x]\]

= cos x/sin x

= cot x

= RHS

Hence proved.

i

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