Question

If sec x+tan x=k, then what is the value of sin x?

Answer 1

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Answer 2

we know that sin²x + cos²x = 1

dividing the above expression with cos²x we get

tan²x + 1 = sec²x

⇒ sec²x - tan²x = 1

⇒ (secx - tanx) (secx + tanx) = 1    ------    since ( a² - b² = (a + b) (a - b))

given secx + tanx = k substituting in the above we get :

⇒ secx - tanx = 1/k

⇒ (secx + tanx) + (secx - tanx) = k + 1/k

⇒ 2secx = (k² + 1/)k

⇒ secx = (k² + 1)/2k

we know that cosx = 1/secx

⇒ cosx = 2k/(k² + 1)

given secx + tanx = k

⇒ 1/cosx + sinx/cosx = k               (since tanx = sinx/cosx)

⇒ 1 + sinx = k × cosx

but we got that cosx =  2k/(k² + 1) substituting we get:

⇒ 1 + sinx = k ×2k/(k² + 1)

⇒ sinx = {2k²/(k² + 1)} - 1

⇒ sinx = (2k² - k² - 1)/(k² + 1)

⇒ sinx = (k² - 1)/(k² + 1)

hence the values of sinx = (k² - 1)/(k² + 1)