Given,
Cosec A = root 10 = H/P
So,
We use Pythagoras theorem, we have
H^2 =P^2 + B^2
1^2 - root 10^2 = B^2
9 = B^2
3 = B

So, sin A = 1/root 10
Cos A= 3/ root 10
Tan A = 1/3
Cot A = 3
Sec A = root 10 /3

Problem: If cosec A=√10, find the value of all t-ratio of theta.

Solution:

To solve the problem, we need to find the values of all six trigonometric ratios of angle A, given cosec A=√10. Let's use the following formulae:

- cosec A = 1/sin A
- sec A = 1/cos A
- tan A = sin A/cos A
- cot A = cos A/sin A
- sin^2 A + cos^2 A = 1
- 1 + tan^2 A = sec^2 A
- 1 + cot^2 A = cosec^2 A

Step 1: Find sin A

cosec A = 1/sin A

√10 = 1/sin A

sin A = 1/√10

sin A = √10/10

Step 2: Find cos A

sin^2 A + cos^2 A = 1

cos^2 A = 1 - sin^2 A

cos^2 A = 1 - (10/100)

cos^2 A = 9/10

cos A = ± √(9/10)

Since A is an acute angle, cos A is positive.

cos A = √(9/10)

cos A = 3/√10

cos A = 3√10/10

Step 3: Find tan A

tan A = sin A/cos A

tan A = (√10/10) / (3√10/10)

tan A = (√10/10) * (10/3√10)

tan A = 1/3

Step 4: Find cot A

cot A = cos A/sin A

cot A = (3√10/10) / (√10/10)

cot A = 3

Step 5: Find sec A

sec A = 1/cos A

sec A = 1 / (3√10/10)

sec A = 10/3√10

Step 6: Find cosec A

cosec A = √10

Step 7: Summarize the values of all six trigonometric ratios

sin A = √10/10

cos A = 3√10/10

tan A = 1/3

cot A = 3

sec A = 10/3√10

cosec A = √10

Therefore, the values of all six trigonometric ratios of angle A are sin A = √10/10, cos A = 3√10/10, tan A = 1/3, cot A = 3, sec A = 10/3√10, and cosec A = √10.