If `f(x)=x^(11)+x^9-x^7+x^3+1` and `f(sin^(-1)(sin8)=alpha,alpha` is constant, then `f(tan^(-1)(t a n8)` is equal to `alpha` (b) `alpha-2` (c) `alpha+

We know `tan theta=cot theta-2 cot 2 theta`. Putting `theta=alpha,2alpha,2^(2)alpha,....................` in (i), we get `tan alpha=(cot alpha-2 cot 2alpha)` `2(tan2alpha)=2(cot 2 alpha-2 cot 2 ^(2)alpha)` `2^(2)(tan2^(2)alpha)=2^(2)(cot2^(2)alpha-2cot2^(3)alpha)` `2^(n-1)(tan2^(n-1)alpha)=2^(n-1)(cot2^(n-1)alpha-2cos2^(n)alpha)` Adding, `tan alpha+2tan...

Correct Answer - C `"sin"(alpha+beta)"sin"(alpha-beta)="sin" gamma(2"sin"beta + "sin"gamma)` `"or ""sin"^(2)alpha-("sin"beta + "sin" gamma)^(2)=0` `"or "("sin"alpha+"sin" beta+"sin"gamma)("sin"alpha-"sin"beta-"sin"gamma)=0` `"Since " 0 lt alpha, beta, gamma lt pi," we have"` `"sin"alpha+"sin"beta+"sin"gamma ne 0` `therefore...

Correct Answer - A We have, `A(alpha, beta)^(-1)=1/e^(beta) [(e^(beta) cos alpha,-e^(beta) sin alpha,0),(e^(beta) sin alpha,e^(beta) cos alpha,0),(0,0,1)]` `=A(-alpha, -beta)`

(i) Since `1 in [-pi//2, pi//2], sin^(-1) (sin 1) = 1` (ii) Since `2 in [-pi//2, pi//2], sin^(-1) (sin 2) != 2` `:. Sin^(-1) (sin 2) = sin^(-1) (sin (pi)...

Correct Answer - (a) `-(2pi)/(5)` (b) `(pi)/(3)` (c) `(2pi)/(5)` (d) `3 pi -8` (e) `10 - 3 pi` (f) `9 - 2pi` (g) `6 - pi` (h) `7 - 2pi` (a)...

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 `x = sin theta and sqrtx = sin phi, " where " x in [0, 1]` `rArr theta, phi in [0, pi//2]` `rArr theta - phi...

Correct Answer - A `(a)` `a=e^(ialpha)`, `b=e^(ibeta)`, `c=e^(igamma)` Clearly `a+b+c=0` `impliesa+b+c=0=(1)/(a)+(1)/(b)+(1)/(c )` `impliesa^(3)+b^(3)+x^(3)=3abc` `impliessin3alpha+sin3beta+sin3gamma=sin(alpha+beta+gamma)`

Answer: (1)  tan1>tan2