Question

10. Find the value of sin 600.cos 2450-Cotet 30 +
5 cos 90°

Answer 1

Step-by-step explanation:

We have,

\dfrac{\sin 60}{\cos ^{2}45} -\cot 30+15\cos 90

cos

2

45

sin60

−cot30+15cos90

Evaluate, \dfrac{\sin 60}{\cos ^{2}45} -\cot 30+15\cos 90

cos

2

45

sin60

−cot30+15cos90 = ?

∴ \dfrac{\sin 60}{\cos ^{2}45} -\cot 30+15\cos 90

cos

2

45

sin60

−cot30+15cos90

=\dfrac{\dfrac{\sqrt{3}}{2} }{(\dfrac{1}{\sqrt{2}})^{2} } -\sqrt{3} +15(0)=

(

2

1

)

2

2

3

−

3

+15(0)

[ ∵ \sin 60=\dfrac{\sqrt{3}}{2} ,\cos 45=\dfrac{1}{\sqrt{2}} ,\cot 30=\sqrt{3} and \cos 90=1sin60=

2

3

,cos45=

2

1

,cot30=

3

andcos90=1

=\dfrac{\dfrac{\sqrt{3}}{2} }{\dfrac{1}{2}} -\sqrt{3} +0=

2

1

2

3

−

3

+0

=\sqrt{3} -\sqrt{3}=

3

−

3

= 0

Hence, \dfrac{\sin 60}{\cos ^{2}45} -\cot 30+15\cos 90=0

cos

2

45

sin60

−cot30+15cos90=0