If #8cosx + 15sinx = 15# and #cosx!=0# then which of the following option is correct for #8sinx-15cosx#?

#a)8#
#b)-8#

2 Answers
Ratnaker Mehta
Jan 13, 2018

#b) -8.#

Explanation:

Given that, #8cosx+15sinx=15#.

#rArr 15sinx=15-8cosx#.

#rArr 225sin^2x=225-240cosx+64cos^2x....[because," squaring]"#.

#rArr 225(1-cos^2x)=225-240cosx+64cos^2x#.

#rArr225-225cos^2x=225-240cosx+64cos^2x#.

#rArr289cos^2x-240cosx=0#

#rArr cosx(289cosx-240)=0#.

#rArr cosx=0, or, cosx=240/289#.

But, we are given that, #cosx ne 0, :. cosx=240/289#.

Utilising this value of #cosx# in #15sinx=15-8cosx#, we get,

#sinx=1/15(15-8cosx)=1-8/15*240/289=1-128/289=161/289#

#:."The Reqd. Val.="8sinx-15cosx#,

#=8*161/289-15*240/289#,

#=(1288-3600)/289#,

#=-2312/289#.

#rArr 8sinx-15cosx=-8#.

Therefore, #b) -8# is the right option.

P dilip_k
Jan 13, 2018

Given

#8cosx+15sinx=15......(1)#

Now

#(8sinx-15cosx)^2+(8cosx+15sinx)^2=8^2(sin^2x+cos^2x)+15^2(cos^2x+sin^2x)-2*8*15sinxcosx+2*8*15sinxcosx#

#=>(8sinx-15cosx)^2+15^2=8^2+15^2#

#=>(8sinx-15cosx)^2=8^2#

#=>8sinx-15cosx=pm8#

So

#8sinx-15cosx=8........(2)#

and

#8sinx-15cosx=-8........(3)#

Now we are to find out which one between (2) and (3) is true.

Adding (1) and (2) we get

#23sinx-7cosx=23# This relation is possible only if #cosx=0# but given condition is #cosx!=0#

So relation (2) is rejected and relation (3) is accepted.

This means option (1) is correct.

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