Question

If cosec a= 7 / 4 then prove that 1 + tan square = secsquare A

Answer 1

Cosec A = 7/4 = H/P
=> H = 7k and P = 4k
Using Pythagoras theorem,

H² = P²+B²

(7k)² = (4k)²+B²

=> B² = 49k²-16k²

=> B = √(33k²) = √(33) k

Now,

LHS:

1+tan²A
= 1+(P/B)²
= 1+(4/√33)²
= 1+(16/33) = 49/33

RHS:

sec²A
= (H/B)²
= (7/√33)² = 49/33

LHS = RHS

Proved

Answer 2

cosec A= 7/4=h/p
base²= hypotenuse² - perpendicular²
=49-16=33
base= √33
1+ tan² A
=1+(4/√33)=1+16/33=49/33=(7/√33)²=(h/b)²=sec²A