\(\frac{{3{\rm{tan}}{{20}^0} - {\rm{ta}}{{\rm{n}}^3}{{20}^0}}}{{1 - 3{\rm{ta}}{{\rm{n}}^3}{{20}^0}}}\) ?
1). \(\frac{1}{{\sqrt 3 }}\)
2). 1
3). \(\sqrt 3 \)
4). \(\infty \)
$(\frac{{3{\rm{tan}}{{20}^0} - {\rm{ta}}{{\rm{n}}^3}{{20}^0}}}{{1 - 3{\rm{ta}}{{\rm{n}}^3}{{20}^0}}})$
[tan3A $( = \frac{{3tanA - ta{n^3}A}}{{1 - 3ta{n^3}A}})$]
tan3 × 20°
∴ tan 60° $( = \sqrt 3 )$
1. By expressing cos 113° in terms of trigonometrical ratios, answer will be
2. If x = sin θ + cos θ and y = sin θ – cos θ, then the value of (x2 + y2 - 1) × (x2 + y2 + 5) will be
3. if $cos\pi x$ = $x^{2} - x +\frac{5}{4}$ , the value of x will be
4. Arc tan [2 cos (arc sin [(3^(1/2))/2]) / 2]) is equal to
7. If $sec\theta + tan\theta$ = $2+\sqrt{5}$ ,then the value of $sin\theta + cos\theta$ is
8. If sinA = cosA and A is acute angle, then tanA - cotA is equal to -
9. Value of (tan1° tan2° tan3° ………..tan89°) is:
10. What is the value of [(sin4x + sin4y) × tan(2x – 2y)] / (sin4x – sin4y)?