`tan(pi/4+1/2cos^-1x)+tan(pi/4-1/2cos^-1x)`, `x!=0` is equal to

`tan^(-1)[2 cos ( 2 sin^(-1)""(1)/(2))]` `=tan^(-1)[2 cos(2 sin^(-1)sin""(pi)/(6))]` `=tan^(-1)[2 cos (2*(pi)/(6))]` `=tan^(-1)(2*(1)/(2))=tan^(-1)(1)` `=tan^(-1)(tan""(pi)/(4))=(pi)/(4)`

LHS=`tan^(-1)""(1)/(5)+tan^(-1)""(1)/(7)+ tan^(-1)""(1)/(3)+tan^(-1)""(1)/(8)` `tan^(-1)""((1)/(5)+(1)/(7))/(1-(1)/(5)xx(1)/(7))+tan^(-1)""((1)/(3)+(1)/(8))/(1-(1)/(3)xx(1)/(8))` `=tan^(-1)""((7+5)/(35))/((35-1)/(35))+tan^(-1)""(11)/(23)` `=tan^(-1)""((6)/(17)+(11)/(23))/(1-(6)/(17)xx(11)/(23))=tan^(-1)""((138+187)/(391))/((391-66)/(391))` `tan^(-1)""(352)/(325)=tan^(-1)1=(pi)/(4)`= RHS Hence Proved

Correct Answer - C `I_(1)int_(0)^(pi//2)(cos^(2)x)/(1+cos^(2)x)dx` `=int_(0)^(pi//2)(cos^(2)(pi//2-x))/(1+cos^(2)(pi//2-x))dx` `=int_(0)^(pi//2) (sin^(2)x)/(1+sin^(2)x)dx=I_(2)` Also `I_(1)+I_(2)=int_(0)^(pi//2)((sin^(2)x)/(1+sin^(2)x)+(cos^(2)x)/(1+cos^(2)x))dx` `=int_(0)^(pi//2)(sin^(2)x+sin^(2)xcos^(2)x+cos^(2)x+sin^(2)xcos^(2)x)/(1+sin^(2)x+cos^(2)x+sin^(2)xcos^(2)x)` `=int(1+2sin^(2)xcos^(2)x)/(2+sin^(2)x cos^(2)x)dx=2I_(3)` `:. 2I_(1)=2I_(3)` or `I_(1)=I_(3)` or `I_(1)=I_(2)=I_(3)`

We have, `[(1,tan theta),(-tan theta,1)]^(-1) =1/(1+ tan^(2) theta) [(1,- tan theta),(tan theta,1)]` `implies [(a,-b),(b,a)]=cos^(2) theta [(1,-tan theta),(tan theta,1)][(1,-tan theta),(tan theta,1)]` `=cos^(2) theta [(1-tan^(2) theta,-2 tan theta),(2 tan theta,1-tan^(2) theta)]` `=[(cos...

Here, `tan^(_1) ((a_(1) x- y)/(a + a_(1) y)) = tan^(-1) ((a_(1) - (y)/(x))/(1 + a_(1) (y)/(x))) = tan^(-1) a_(1) - tan^(-1) (y)/(x)` `tan^(-1) ((a_(2) -a_(1))/(1 + a_(2) a_(1))) = tan^(-1)...

(i) `tan^(-1) (tan.(2pi)/(3)) != (2pi)/(3), " as " (2pi)/(3)` does not lie between `- (pi)/(2) and (pi)/(2)` Now, `tan^(-1) (tan.(2pi)/(3)) = tan^(-1) (tan(pi - (pi)/(3)))` `= tan^(-1) (- tan. (pi)/(3))`...

Correct Answer - C `2 tan^(-1 (cosec tan^(-1) x - tan cot^(-1) x)` `= 2 tan^(-1) [cosec {cosec^(-1) (sqrt(1 + x^(2)))/(x)} - tan^(-1) {tan^(-1) ((1)/(x))}]` `= 2 tan^(-1) [sqrt((1 + x^(2))/(x))...

Correct Answer - A Let `sqrt(tan alpha) = tan x`. Then `u = cot^(-1) (tan x) - tan^(-1) (tan x)` `= (pi)/(2) - x -x = (pi)/(2) - 2x` or `2x...

Correct Answer - A `theta = tan^(-1) (2 tan^(2) theta) - tan^(-1) ((1)/(3) tan theta) = tan^(-1).(2 tan^(2) theta - (1)/(3) tan theta)/(1 + (1)/(3) tan^(3) theta)` `rArr tan theta =...

Correct Answer - B Let us find the value of general term of the series `S = 7 + 19 + 39 + 67 + .. + T_(n)` `(S = "...