If 3 sin A = 4 cos A ; find the value of
(i) sec square A
(ii) sin A / cosec A + cos A / sec A
Answers
Answer:
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
Answer:
1)25/9
2)1
Step-by-step explanation:
Given:-
3 sin A = 4 cos A
To find:-
Find the value of
ind the value of (i) sec square A
ind the value of (i) sec square A(ii) sin A / cosec A + cos A / sec A.
Solution:-
3 Sin A= 4 Cos A
=> Sin A/Cos A=4/3
=>Tan A=4/3
On squaring both sides ,then
=>(Tan A)²=(4/3)²
=>Tan² A=16/9
On adding 1 both sides then
=>1+Tan² A=1+(16/9)
=>Sec²A=(9+16)/9
(1)Sec² A=25/9
=>Sec A=√(25/9)
=>Sec A=5/3
=>1/Cos A=5/3
=>Cos A=3/5
=>Cos²A=9/25
=>1-Cos² A=1-(9/25)
=>Sin²A=(25-9)/25
=>Sin²A=16/25
=>Sin A=√(16/25)
=>Sin A=4/5
=>1/Sin A=5/4
=>Cosec A=5/4
2)sin A / cosec A + cos A / sec A
=>[(4/5)/(5/4)]+[(3/5)/(5/3)]
=>[(4/5)×(4/5)]+[(3/5)×(3/5)]
=>(16/25)+(9/25)
=>(16+9)/25
=>25/25
=>1
sin A / cosec A + cos A / sec A =1
(or)
sin A / cosec A + cos A / sec A
=>sin A×Sin A + Cos A×CosA
=>Sin² A+Cos² A
=>1
sin A / cosec A + cos A / sec A =1