Question

Sin80-cos70 iska hal chahiye

Answer 1

We have to prove that sin80^{{\circ}}-cos70^{{\circ}}=cos50^{{\circ}}sin80

∘

−cos70

∘

=cos50

∘

⇒sin80^{{\circ}}=cos50^{{\circ}}+cos70^{{\circ}}sin80

∘

=cos50

∘

+cos70

∘

Now, using formula, cos(C+D)= 2cos\frac{C+D}{2}cos\frac{C-D}{2}cos(C+D)=2cos

2

C+D

cos

2

C−D

⇒sin80^{{\circ}}=2 cos\frac{50+70}{2}cos\frac{50-70}{2}sin80

∘

=2cos

2

50+70

cos

2

50−70

⇒sin80^{{\circ}}=2cos60^{{\circ}}cos(-10)^{{\circ}}sin80

∘

=2cos60

∘

cos(−10)

∘

We know that,cos(-\alpha)=cos\alphacos(−α)=cosα ,therefore,

⇒sin80^{{\circ}}=2cos60^{{\circ}}cos10^{{\circ}}sin80

∘

=2cos60

∘

cos10

∘

⇒sin80^{{\circ}}=2{\times}\frac{1}{2}{\times}cos(90^{{\circ}}-80^{{\circ}})sin80

∘

=2×

2

1

×cos(90

∘

−80

∘

)

⇒sin80^{{\circ}}=sin80^{{\circ}}sin80

∘

=sin80

∘

Since, L.H.S=R.H.S, hence proved.