`theta = tan^(-1) (2 tan^(2) theta) - tan^(-1) ((1)/(3) tan theta) " then " tan theta=`

Correct Answer - `2//e` `I_(2)int_(0)^(pi//4)(e^(tan^(2)theta).tan theta)/((2-tan^(2)theta))sec^(2) d theta` Put `tan^(2)theta=t` `:.2 tan theta sec^(2) d theta =dt` `:.I_(2)=1/2int_(0)^(1)(e^(t)dt)/((2-t))` `=1/2 int_(0)^(1)(e^(1-t)dt)/(1+t)` `=e/2int_(0)^(1)(dt)/(e^(t)(t+1))=e/2.I_(1)` `:. (I_(1))/(I_(2))=2/e`

Correct Answer - A Putting `x tan theta =z sin theta` or `dx=cos theta dz`. We get `I=cos theta int_(tan theta)^(1) f(z sin theta)dz` `=-cos theta int_(1)^(tan theta)f(x sin theta) dx`

By defing A & B are equal if they have the same order and all the corresponding elements are equal. Thus we have `sin theta=(1)/(sqrt2),c os theta=-(1)/(sqrt2)& tan theta=-1` `Rightarrow...

Let matrix `A=[(a,b),(c,d)]` is orthogonal matrix. `:. [(a,b),(c,d)][(a,c),(b,d)]=[(1,0),(0,1)]` `implies [(a^(2)+b^(2),ac+bd),(ac+bd,c^(2)+d^(2))]=[(1,0),(0,1)]` `implies a^(2)+b^(2)=1` (1) `c^(2)+d^(2)=1` (2) `ac+bd=0` (3) `implies a/d= (-b)/c= k` (let) `implies c^(2)+d^(2)=(a^(2)+b^(2))/k=1//k^(2)` or `k^(2)=1` or `k= pm 1`...

Correct Answer - C `AB=[(cos^(2) theta,),(cos theta sin theta,)][(cos^(2) phi,cos phi sin phi),(cos phi sin phi,sin^(2) phi)]` `=[(cos^(2) theta cos^(2) phi+cos theta cos phi sin theta sin phi ,cos^(2)theta cos phi...

Correct Answer - A::B::C We have, `|A(theta)|=1` Hence, A is invertiable. `A(pi+theta)=A(pi + theta)=[(sin(pi+theta),i cos (pi+theta)),(i cos (pi+theta),sin (pi+theta))]` `=[(-sin theta,-i co theta),(-i cos theta,-sin theta)]=-A(theta)` adj `(A(theta))=[(sin theta,-i cos theta),(-i...

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...

Correct Answer - 9 We must have `4 sin^(2) theta + sin theta = 6 sin theta -1` `rArr 4 sin^(2) theta - 5 sin theta + 1 = 0` `rArr...

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 - `[{:(,1,0),(,1,1):}]`