Question

3. If cosec 0 = V10 then sec 0 =?​

Answer 1

Answer:

Given:-

cosec A=√10

Solution

cosecA=√10

H/P=√10

H=√10x units

P= 1x units

According to Pythagoras theorem:-

$P^2+B^2=H^2$

${x}^{2} + B^2 = { \sqrt{10} x}^{2}$

${x}^{2} + B^2 = 10 {x}^{2}$

$B^2 = {10x}^{2} - {x}^{2}$

$B^2 = {9x}^{2}$

$B = \sqrt{9 {x}^{2} }$

B=3x units

Now ,

Since,

sec A=H/B

$= \frac{ \sqrt{10} }{3}$

Answer 2

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