Introduction to...

If $$\sec x=p+\dfrac 1{4p}$$, then $$\sec x+\tan x$$

$$\sin 30^{o}- cos60^{o}$$= ?

In the given figure, $$\angle PAB = 60^o $$ and $$ \angle CAB = 30^o $$ and $$PB = 75m.$$ Find $$BC.$$

Solve $$\sin A(1+\tan A)+\cos A(1+\cot A)$$.

If $$\tan A+\cot A=4,\ then\ {\tan}^{4}\ A+{\cot}^{4}\ A$$ is equal to 

If $$x\sin \,{45^ \circ }{\cos ^2}{60^ \circ } = \dfrac{{{{\tan }^2}{{60}^ \circ }\cos ec{{30}^ \circ }}}{{\sec {{45}^ \circ }{{\cot }^2}{{30}^ \circ }}}$$ then x =

If $$x = a \cos \theta \text { and } y = b \sin \theta$$, then $$b ^ { 2 } x ^ { 2 } + a ^ { 2 } y ^ { 2 }$$ is equal to 

If $$\cos 9\theta=\sin \dfrac{\pi}{4}$$, then the value of $$\tan 6\theta$$ is

If $$\tan { x } =3\cot { x } ,$$ then x =?

$$\cos 35 ^ { \circ } + \cos 85 ^ { \circ } + \cos 155 ^ { \circ }$$ is