If f(x) =tan , `x in [0.(pi)/(5)]` then show that `(pi)/(5) lt ((pi)/(5)) lt (2pi)/(5)`

Find the value tan (pi/5)+2tan((2pi)/5)+4cot((4pi)/5).

For x in (0,(pi)/(2)) prove that "sin"^(2) x lt x^(2) lt " tan"^(2) x

If f(x)= sin (arctan x), g(x) = tan (arcsin x), and 0 le x lt (pi)/(2) , then f(g((pi)/(10)))=

which of the following statement is // are true ? (i) f(x) =sin x is increasing in interval [(-pi)/(2),(pi)/(2)] (ii) f(x) = sin x is increasing at all point of the interval [(-pi)/(2),(pi)/(2)] (3) f(x) = sin x is increasing in interval ((-pi)/(2),(pi)/(2)) UU ((3pi)/(2),(5pi)/(2)) (4) f(x)=sin x is increasing at all point of the interval ((-pi)/(2),(pi)/(2)) UU ((3pi)/(2),(5pi)/(2)) (5) f(x) = sin x is increasing in intervals [(-pi)/(2),(pi)/(2)]& [(3pi)/(2),(5pi)/(2)]

If tan x= (3)/(4) " where " pi lt x lt (3pi)/( 2), value of tan ""(x)/(2) is

Let f(x) ={{:( x+2, 0 le x lt 2),( 6-x, x ge 2):}, g(x) ={{:( 1+ tan x, 0le x lt (pi) /(4)),( 3-cotx,(pi)/(4) le x lt pi ):} f(g(x)) is

Lt_(x to (pi)/(2))(pi/2-x)tan x is equal to (i) 1 (ii) -1 (iii) (pi)/(2) (iv) (2)/(pi)

Simplify [(3 sin 2 alpha)/(5 + 3 cos 2 alpha)] + tan^(-1) [(tan alpha)/(4)], " where " -(pi)/(2) lt alpha lt (pi)/(2)

If sec x cos 5x=-1 and 0 lt x lt (pi)/(4) , then x is equal to

-1lt cos(2x -(pi)/(3))lt(1)/(2) has solution set (A) x in ((pi)/(3)+2n pi, (5pi)/(3)+2n pi), n in I (B) x in ((pi)/(3)+2n pi, (2pi)/(3)+2n pi)uu((2pi)/(3)+2n pi, (2n+1)pi), n in I (C) x in ((pi)/(3)+2n pi, pi+2n pi)uu(pi+2n pi, (5pi)/(3)+2n pi), n in I (D) x in ((pi)/(3)+n pi, (2pi)/(3)+n pi)uu((2pi)/(3)+n pi, (n+1)pi), n in I