If `(1-t a nthetat a ntheta1)(1t a ntheta-t a ntheta1)=[a-bb a]` , then `a=cos2theta` b. `a=1` c. `b=s in2theta` d. `b=-1`

If [1-tanthetatantheta1][1tantheta-tantheta1]^(-1)=[a-b b a] , then a=1,\ \ b=1 (b) a=cos2\ theta,\ \ b=sin2\ theta (c) a=sin2\ theta,\ \ b=cos2theta (d) none of these

If ((1,-tantheta), (tantheta,1))((1,tantheta), (-tantheta,1))^(-1)=[(a,-b), (b,a)], then (i) a=cos2theta (ii) b=1 (iii) b=sin2theta (iv) b=-1

If A=[(1,-tantheta),(tantheta, 1)] [(1,tantheta),(-tantheta,1)]^-1 = [(a,-b),(b,a)] then (A) a=b=-1 (B) a=sin2theta, b=cos2theta (C) a=cos2theta, b=sin2theta (D) none of these

Evaluate: intsqrt(t a ntheta)\ d theta

(csc theta - sin theta)(sec theta - cos theta)(tan theta+ cot theta)= A)1 B)0 C) -1 D) 2

If ((1 + cos theta + i sin theta ) /(i+ sin theta + i cos theta ))= cos ntheta + i sin n theta then n is equal to :

Prove that: a) (sin2theta)/(1+cos2theta) = tantheta b) (1+sin2theta+cos2theta)/(1+sin2theta-cos2theta)=cottheta

Prove that (frac(1+costheta+isintheta)(1+costheta-isintheta))^n=cos ntheta+isin ntheta .

a cos theta+b sin theta+c=0a cos theta+b_(1)sin theta+c_(1)=0