Trigonometry Quiz 11

$\sqrt3$ Tan $\theta$ = 1, then $\theta$ =

In a $\triangle ABC$ if $\angle B$ = 90° and Tan C = $\frac{5}{12}$ , then the length of the hypotenuse is

$Tan^2 60^\circ + 4 Cos^2 45^\circ + 3 Sec^2 30^\circ + 5 Cos^2 90^\circ=$

$Sec^2 27^\circ-Cot^263^\circ=$

$\sqrt{\frac{1}{Sec^2 \theta} + Sin^2 \theta + \frac{1}{Tan^2 \theta}} =$

$Sin^2 \theta (1 + Cot^2 \theta) =$

If A = 45°, then $(1 + Tan A) (1 + Tan^2 A) (1 + Tan^3 A)$ =

$Sin^2 \theta . Cot^2 \theta + Cos^2 \theta . Tan^2 \theta =$

$Tan\,45^\circ+2\,Tan^260^\circ=$

Sin 2A = 2 Sin A, then $\angle A$ =