3. If cosec 0 = V10 then sec 0 =?
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Answer:
Given:-
cosec A=√10
Solution
cosecA=√10
H/P=√10
H=√10x units
P= 1x units
According to Pythagoras theorem:-
[tex]P^2+B^2=H^2[/tex]
[tex] {x}^{2} + B^2 = { \sqrt{10} x}^{2} [/tex]
[tex] {x}^{2} + B^2 = 10 {x}^{2} [/tex]
[tex]B^2 = {10x}^{2} - {x}^{2} [/tex]
[tex]B^2 = {9x}^{2} [/tex]
[tex]B = \sqrt{9 {x}^{2} } [/tex]
B=3x units
Now ,
Since,
sec A=H/B
[tex] = \frac{ \sqrt{10} }{3} [/tex]
Given:-
cosec A=√10
Solution
cosecA=√10
H/P=√10
H=√10x units
P= 1x units
According to Pythagoras theorem:-
P^2+B^2=H^2P
2
+B
2
=H
2
{x}^{2} + B^2 = { \sqrt{10} x}^{2}x
2
+B
2
=
10
x
2
{x}^{2} + B^2 = 10 {x}^{2}x
2
+B
2
=10x
2
B^2 = {10x}^{2} - {x}^{2}B
2
=10x
2
−x
2
B^2 = {9x}^{2}B
2
=9x
2
B = \sqrt{9 {x}^{2} }B=
9x
2
B=3x units
Now ,
Since,
sec A=H/B
= \frac{ \sqrt{10} }{3}=
3
10