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Home » Prove that: (i) tan 8x – tan 6x – tan 2x = tan 8x tan 6x tan 2x (ii) tan π/12 + tan π/6 + tan π/12 tan π/6 = 1
Prove that: (i) tan 8x – tan 6x – tan 2x = tan 8x tan 6x tan 2x (ii) tan π/12 + tan π/6 + tan π/12 tan π/6 = 1
CBSE | Class 11 | Exercise 7.1 | Maths | RD Sharma | Values of Trigonometric Functions at Sum or Difference of Angles
Prove that: (i) tan 8x – tan 6x – tan 2x = tan 8x tan 6x tan 2x (ii) tan π/12 + tan π/6 + tan π/12 tan π/6 = 1

    \[\left( \mathbf{i} \right)tan\text{ }8x\text{ }-\text{ }tan\text{ }6x\text{ }-\text{ }tan\text{ }2x\text{ }=\text{}tan\text{}8x\text{}tan\text{ }6x\text{ }tan\text{ }2x\]

Let us consider LHS:

    \[tan\text{ }8x\text{ }\text{ }tan\text{ }6x\text{ }\text{ }tan\text{ }2x\]

    \[tan\text{ }8x\text{ }=\text{ }tan\left( 6x\text{ }+\text{ }2x \right)\]

Since,

    \[tan\text{ }\left( A\text{ }+\text{ }B \right)\text{ }=\text{ }\left(tan\text{}A\text{}+\text{}tan\text{}B\right)\text{}\text{ }\left( 1\text{}-\text{}tan\text{}A\text{}tan\text{}B\right)\]

Therefore,

    \[tan\text{ }8x\text{ }=\text{ }\left( tan\text{ }6x\text{ }+\text{}tan\text{}2x\right)\text{}/\text{}\left(1\text{}-\text{}tan\text{}6x\text{}tan\text{}2x\right)\]

By cross-multiplying we get,

    \[tan\text{ }8x\text{ }\left( 1\text{ }-\text{ }tan\text{ }6x\text{ }tan\text{ }2x \right)\text{ }=\text{ }tan\text{ }6x\text{ }+\text{ }tan\text{ }2x\]

    \[tan\text{ }8x\text{}-\text{}tan\text{}8x\text{}tan\text{}6x\text{}tan2x\text{}=\text{ }tan\text{ }6x\text{ }+\text{ }tan\text{ }2x\]

Rearranging we get,

    \[tan\text{}8x\text{ }-\text{ }tan\text{ }6x\text{ }-\text{ }tan\text{ }2x\text{ }=\text{}tan\text{}8x\text{}tan\text{}6x\text{}tan\text{ }2x\]

    \[=\text{ }RHS\]

    \[\therefore LHS\text{ }=\text{ }RHS\]

Hence proved.

(ii) 

    \[tan\text{ }\pi /12\text{ }+\text{ }tan\text{ }\pi /6\text{ }+\text{ }tan\text{ }\pi /12\text{ }tan\text{ }\pi /6\text{ }=\text{ }1\]

We know,

    \[\pi /12\text{ }=\text{ }15{}^\circ \text{ }and\text{ }\pi /6\text{ }=\text{ }30{}^\circ \]

Therefore, we have

    \[15{}^\circ \text{ }+\text{ }30{}^\circ \text{ }=\text{ }45{}^\circ \]

    \[Tan\text{ }\left( 15{}^\circ \text{ }+\text{ }30{}^\circ \right)\text{ }=\text{ }tan\text{ }45{}^\circ \]

Since,

    \[tan\text{}\left(A\text{}+\text{}B\right)\text{}=\text{}\left(tan\text{}A\text{}+\text{}tan\text{}B\right)\text{}/\text{}\left(1\text{}-\text{}tan\text{}A\text{}tan\text{}B\right)\]

Therefore,

    \[(tan\text{}{{15}^{o}}+\text{}tan\text{}{{30}^{o}})\text{}/\text{}(1\text{}-\text{}tan\text{}{{15}^{o}}tan\text{}{{30}^{o}})\text{}=\text{ }1\]

    \[tan\text{}15{}^\circ \text{}+\text{}tan\text{}30{}^\circ\text{}=\text{}1\text{}-\text{}tan\text{}15{}^\circ\text{}tan\text{}30{}^\circ\]

Upon rearranging we get,

    \[tan15{}^\circ \text{ }+\text{ }tan30{}^\circ \text{ }+\text{ }tan15{}^\circ \text{ }tan30{}^\circ \text{ }=\text{ }1\]

Hence proved.

i

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