12. Given : 4 sin = 3 cos 0; find the value of -
(i) sin 0
(ii) cos 0
(iii) cot2 0 - cosec 0
(iv) 4 cos2 0 - 3 sin2 0 + 2
Answers
Answered by
1
Answer:
Correct option is
A
4
3cosA=4sinA
tanA=
4
3
tanA=
B
P
=
4
3
Using Pythagoras Theorem,
H
2
=P
2
+B
2
H
2
=3
2
+4
2
H
2
=5
2
H=5
Now, 3−cot
2
A+csc
2
A = 3−(
P
B
)
2
+(
P
H
)
2
= 3−(
3
4
)
2
+(
3
5
)
2
= 3−
9
16
+
9
25
=
9
27−16+25
=
9
36
= 4
answers:- (iv)4cos2 0-3sin2 0+2
Answered by
2
Step-by-step explanation:
4Sin=3 cos theta
sin theta = 3 cos theta/4
Sin theta /Cos theta =3/4
Tan theta = 3/4
Sin theta = 3
cos Theta = 4
Cot theta = 4/3
cosec theta = 1/3
iii.Cot2 thetha - cosec theta
(4/3)2- 1/3
16/9-1/3
16-3
____
9
13/9
iv.4(4)2 - 3(3)2+ 2
4*16-3*9+2
64-27+3
64-60
4
Hope it helps....
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