[" 17.aft sin "theta=(p)/(sqrt(p^(2)+q^(2)))" ,at fere affere fas "],[qquad q sin theta-p cos theta]

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

5. sin p theta+sin q theta=0

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

If tan theta=(p)/(q) then the value of (p sin theta-q cos theta)/(p sin theta+q cos theta) is

If acos theta+b sin theta=p and a sin theta-b cos theta=q then

If sin theta=(p)/(q), then the value of cos theta+sin 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))

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)=

Let P = {theta , sin theta - cos theta = sqrt(2)cos theta} and Q = {theta , sin theta + cos theta = sqrt(2) sin theta) be two sets. Then