Given, `cos theta=(3)/(5)`
Let PQR be the right triangle such that `angleQPR=theta`( see fig. 16.5)
Assume that PQ = 3 and PR=5 .
Then, `QR=sqrt(PR^(2)=PQ^(2))=sqrt(25-9)=4`
So, `tan theta=("Opposite side ")/("Adjacent side ")=(QR)/(PQ)=(4)/(3) and "cosec" theta=("Hypotenuse")/("Opposite side ")=(PR)/(QR)=(5)/(4)`.