Prove that 3 tan 2 45-sin 2 60-1-2 2 30-1-8 2 45-1

LHS = 3(1)^2 (√(3)/2)^2 1/2(√(3))^2 + 1/8(√(2))^2 = 3 3/4 3/2+1/8(2) = 3 3/4 3/2+1/4 = 12 3 6+1/4 = 1 ...

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Standard Values of Trigonometric Ratios

Prove that 3 ...

Question

Prove that $3 t a n^{2} 45^{\circ} - s i n^{2} 60^{\circ} - (\frac{1}{2}) c o t^{2} 30^{\circ} + (\frac{1}{8}) s e c^{2} 45^{\circ}$ = 1

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{sin}^{2} 72^{\circ} - {sin}^{2} 60^{\circ} = \frac{\sqrt{5} - 1}{8}

Prove that:

$s i n^{2} 72^{\circ} - s i n^{2} 60^{\circ} = \frac{\sqrt{5} - 1}{8}$

s i n^{2} 60^{\circ} + c o s^{2} (3 x - 9^{\circ}) = 1

Prove

sin²72° - sin²60° = √5 -1/8

\sqrt{\frac{1 - s i n^{2} 60^{\circ}}{1 - c o s^{2} 60^{\circ}}}

\frac{1}{\sqrt{m}}

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