If tanθ = \(\frac{12}{5}\) , find the value of \(\frac{1+sinθ}{1-s

1.	If tanθ = \(\frac{12}{5}\) , find the value of \(\frac{1+sinθ}{1-sinθ}\)
Answer» \(\frac{1+sinθ}{1-sinθ}\) = \(\frac{1+sinθ}{1-sinθ}\times \frac{1+sinθ}{1+sinθ}\) = \(\frac{(1+sinθ)^2}{1-sin^2θ}\) = \(\frac{(1+sinθ)^2}{cos^2θ}\) = sec²θ + tan²θ = tan²θ + 1 + tan²θ = 2 tan²θ + 1 = 2 x \(\Big(\frac{12}{5}\Big)^2+1\) \(\Big(∵tanθ=\frac{12}{5}\Big)\) = \(\frac{288}{25}+1\) = \(\frac{288+25}{25}\) = \(\frac{313}{25}\)

If tanθ = \(\frac{12}{5}\) , find the value of \(\frac{1+sinθ}{1-sinθ}\)

Answer»

\(\frac{1+sinθ}{1-sinθ}\) = \(\frac{1+sinθ}{1-sinθ}\times \frac{1+sinθ}{1+sinθ}\)

= \(\frac{(1+sinθ)^2}{1-sin^2θ}\)

= \(\frac{(1+sinθ)^2}{cos^2θ}\)

= sec²θ + tan²θ

= tan²θ + 1 + tan²θ

= 2 tan²θ + 1

= 2 x \(\Big(\frac{12}{5}\Big)^2+1\) \(\Big(∵tanθ=\frac{12}{5}\Big)\)

= \(\frac{288}{25}+1\)

= \(\frac{288+25}{25}\) = \(\frac{313}{25}\)