We have,
3cosθ+4sinθ=92
6cosθ+8sinθ=9
Let αandβ are the different valueθ
Then,
3cosα+4sinα=92
6cosα+8sinα=9......(1)
6cosβ+8sinβ=9......(2)
Subtract equation (2) from (1) and we get,
⇒6(cosα−cosβ)+8(sinα−sinβ)=0
⇒6(cosα−cosβ)=−8(sinα−sinβ)
⇒3(cosβ−cosα)=4(sinα−sinβ)
⇒(sinα−sinβ)(cosβ−cosα)=34
⇒2cos(α+β2)sin(α−β2)2sin(α+β2)sin(α−β2)=34
⇒cos(α+β2)sin(α+β2)=34
⇒cot(α+β2)=34
Squaring both side and we get,
⇒cot2(α+β2)=(34)2
⇒csc2(α+β2)−1=(34)2∴csc2θ−cot2θ=1
⇒csc2(α+β2)=916+1
⇒csc2(α+β2)=2516
⇒sin2(α+β2)=1625
⇒sin(α+β2)=45
But we know that,
sin2θ+cos2θ=1
cos2θ=1−sin2θ
cosθ=√1−sin2θ
cos(α+β2)=√1−sin2(α+β2)
=√1−1625
=√925
cos(α+β2)=35
Now,
sin(α+β)=2sin(α+β2)cos(α+β2)
=2×45×35
=2425
sin(α+β)=2425
Hence, this is the
answer.