sin 35° cos 55° + cos 35° sin 55° / cosec 2-10- tan 280
Answers
Step-by-step explanation:
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Answer:
We have,
\dfrac{\sin 35\cos 55 +\cos 35\sin 55}{\csc^2 10-\tan^2 80}
csc
2
10−tan
2
80
sin35cos55+cos35sin55
To find, \dfrac{\sin 35\cos 55 +\cos 35\sin 55}{\csc^2 10-\tan^2 80}=?
csc
2
10−tan
2
80
sin35cos55+cos35sin55
=?
∴ \dfrac{\sin 35\cos 55 +\cos 35\sin 55}{\csc^2 10-\tan^2 80}
csc
2
10−tan
2
80
sin35cos55+cos35sin55
=\dfrac{\sin (35+55)}{\csc^2 10-\tan^2 (90-10)}=
csc
2
10−tan
2
(90−10)
sin(35+55)
[ ∵ \sin (A+B)=\sin A\cos B +\cos A\sin Bsin(A+B)=sinAcosB+cosAsinB
=\dfrac{\sin 90}{\csc^2 10-\cot^2 10}=
csc
2
10−cot
2
10
sin90
=\dfrac{1}{1}=
1
1
[ ∵ \sin 90=1sin90=1 and \csc^2 A-\cot^2 A=1csc
2
A−cot
2
A=1
= 1
Hence, \dfrac{\sin 35\cos 55 +\cos 35\sin 55}{\csc^2 10-\tan^2 80}=1
csc
2
10−tan
2
80
sin35cos55+cos35sin55
=1