Question

If tanq=3/4 then find the value of secQ and cosQ​

Answer 1

Answer:

we know sec²x - tan²x = 1

so, sec²Q - tan²Q = 1

or, (secQ - tanQ)(secQ - tanQ) = 1

or, (secQ - tanQ) = 1/(secQ + tanQ) = 1/P

now, solve equations ; secQ + tanQ = P and secQ - tanQ = 1/P

e.g., (secQ - tanQ) + (secQ + tanQ) = p + 1/p

2secQ = (p² + 1)/p

secQ = (p² + 1)/2p

cosQ = 2p/(p² + 1) = base/hypotenuse

perpendicular = \sqrt{(p^2+1)^2-(2p)^2}

(p

2

+1)

2

−(2p)

2

= ±(p² - 1)

so, cosecQ = hypotenuse/perpendicular

= ± (p² + 1)/(p² - 1)

Answer 2

Answer:

if tanQ=3/4 =Prepindicular/Base

H²= P² + B²

H²= 3² +4²

H²=9+16

H²=25

H= 5

Then, SecQ= H/B

= 5/4

CosQ=4/5