Question

If 3 cot `theta = 2, `show  that `((4 sin theta - 4 cos theta))/((2 sin theta + 6 cos theta ))=1/3`

 

Answer 1

It is given that cos 𝜃 = `2/3`

LHS = `( 4 sin theta -3 cos theta)/(2 sin theta + 6 cos theta)`

Dividing the above expression by sin 𝜃, 𝑤𝑒 𝑔𝑒𝑡:

`(4-3 cot theta)/(2+6 cot theta)           [∵ cot theta = (cos theta)/(sin theta)]`

Now, substituting the values of cot 𝜃 𝑖𝑛 𝑡ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 𝑒𝑥𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛, 𝑤𝑒 𝑔𝑒𝑡:

`(4-3(2/3))/(2+6(2/3))`

=`(4-2)/(2+4)=2/6=1/3`

i.e., LHS = RHS
Hence proved.