Question

if sec theta + tab theta = p, then find the value of tan theta.​

Answer 1

Answer:

ecθ+tanθ=p ----------------------(1)

∵, sec²θ-tan²θ=1

or, (secθ+tanθ)(secθ-tanθ)=1

or, secθ-tanθ=1/p ----------------(2)

Adding (1) and (2) we get,

2secθ=p+1/p

or, secθ=(p²+1)/2p

∴, cosθ=1/secθ=2p/(p²+1)

∴, sinθ=√(1-cos²θ)

=√[1-{2p/(p²+1)}²]

=√[1-4p²/(p²+1)²]

=√[{(p²+1)²-4p²}/(p²+1)²]

=√[(p⁴+2p²+1-4p²)/(p²+1)²]

=√(p⁴-2p²+1)/(p²+1)

=√(p²-1)²/(p²+1)

=(p²-1)/(p²+1)

∴, cosecθ=1/sinθ=1/[(p²-1)/(p²+1)]=(p²+1)/(p²-1) Ans.

Answer 2

Answer:

ecθ+tanθ=p ----------------------(1)∵,

sec²θ-tan²θ=1or,

(secθ+tanθ)(secθ-tanθ)=1or,

secθ-tanθ=1/p ----------------

(2)Adding (1) and (2) we get,2secθ=p+1/por,

secθ=(p²+1)/2p∴,

cosθ=1/secθ=2p/(p²+1)∴,

sinθ=√(1-cos²θ)=

√[1-{2p/(p²+1)}²]=

√[1-4p²/(p²+1)²]=

√[{(p²+1)²-4p²}/(p²+1)²]=

√[(p⁴+2p²+1-4p²)/(p²+1)²]=

√(p⁴-2p²+1)/(p²+1)=√(p²-1)²/(p²+1)=

(p²-1)/(p²+1)∴,

cosecθ=1/sinθ=1/[(p²-1)/(p²+1)]=(p²+1)/(p²-1)