sin600=1+tan2300tan300.

As we know that sin2A=1+tan2A2tanAPut A=30∘⟹sin2(30∘)=1+tan230∘2tan30∘⟹sin60∘=1+tan230∘2tan30∘So the Given relation is False

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sin60 ^0={tan30^0}{1+tan^{2}30^0}.

Question

$s in 6 0^{0} = \frac{t an 3 0 ^{0}}{1 + t a n ^{2} 3 0 ^{0}}$ .

True

False

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Michael Discussed

s in 6 0^{0} = \frac{t an 3 0 ^{0}}{1 + t a n ^{2} 3 0 ^{0}}

9 mins ago

Discuss this question LIVE

9 mins ago

As we know that

sin 2 A = \frac{2 tan A}{1 + tan ^{2} A}

Put

A = 3 0^{\circ}

⟹ sin 2 (3 0^{\circ}) = \frac{2 tan 3 0 ^{\circ}}{1 + tan ^{2} 3 0 ^{\circ}}

⟹ sin 6 0^{\circ} = \frac{2 tan 3 0 ^{\circ}}{1 + tan ^{2} 3 0 ^{\circ}}

So the Given relation is

False

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