Question

If tanA=1/3 prove that cosec square A=1+ cos squareA

Answer 1

tanA = 1/3 = p/b

from Pythagoras theorem,

h = √(p² + b²) = √(1² + 3²) = √(10)

cosecA = h/p = √(10)/1 ........(1)

cosA = b/h = 3/√(10) ......(2)

now, LHS = cosec²A = (√10)² [ from equation (1),

= 10

RHS = 1 + cos²A = 1 + {3/√(10)}² [ from equation (2),

= 1 + 9/10

= (1 + 9)/10 = 10/10 = 1

LHS ≠ RHS

so, can't possible cosec²A = 1 + cos²A

I think your question is ---> prove that cosec²A = 1 + cot²A

then, LHS = cosec²A = {√(10)}² = 10

cotA = b/p = 3/1 = 3 ......(3)

then, RHS = 1 + cot²A

= 1 + 3² = 10 [ from equation (3),

LHS = RHS ,hence proved

Answer 2

Answer:

Hope this will definitely help you☝