If `sin theta= (p)/(sqrt(p^(2)+q^(2)))` then prove that `qsin theta=p cos theta`

if : p*cos theta=sqrt(p^(2)+q^(2)).sin theta, "then" : q * sin theta=

If tan theta=(p)/(q) then (p sin theta-q cos theta)/(p sin theta+q cos theta)=

If costheta + sin theta = sqrt2 cos theta , prove that cos theta - sin theta = sqrt2 sin theta .

If tan theta=(p)/(q) then (p sin theta-q cos theta)/(p sin tehta+q cos theta)=

Solve : cos p theta = sin q theta.

If tan theta = p/q then (p sin theta - q cos theta)/(p sin theta + q cos theta)=

If tan theta = p/q then (p sin theta - q cos theta)/(p sin theta + q cos theta)=

ABCD is a trapezium such that AB and CD are parallel and BC bot CD . If angleADB = theta, BC = p and CD = q , then AB is equal to (a) ((p^(2)+q^(2))sin theta)/(p cos theta +q sin theta) (b) (p ^(2) + q ^(2)cos theta)/(p cos theta +q sin theta) (c) (p ^(2)+ q^(2))/(p^(2)cos theta +q^(2) sin theta) (d) ((p^(2) +q^(2))sin theta)/((p cos theta + sin theta)^(2))

Let P= {theta : sin theta - cos theta = sqrt2 cos theta} and Q = {theta : sin theta + cos theta = sqrt2 sin theta} be two sets. Then

Let P = {theta: sin theta- cos theta= sqrt2 cos theta} and Q= {theta: sin theta+ cos theta= sqrt2 sin theta} be two sets. Then :