Question

If sec alpha + tan alpha = x then the possible value of sin alpha is
a.1/2(x + 1/x)
b.1/2(x - 1/x)
c.(x*2 - 1)/(x*2 + 1)
d.Can't be determined

Answer 1

Answer:

none of these

Step-by-step explanation:

Given

secα + tanα =x                                                  → Eq1

sec²α - tan²α = 1                                                -it is a trigonometric identity

(secα + tanα) (secα - tanα) =1                            : identity : a²-b² = (a+b)(a-b)

x (secα - tanα) = 1

secα - tanα = 1/x                                               →Eq 2

→Adding eq1 and 2 we get,

2secα = x + 1/x

 secα= 1/2 ( x + 1/x)                                                    → Eq3

→ Subtracting Eq 1 and 2, we get  

2tanα = x - 1/x

tanα = 1/2 (x -1/x)

sinα/cosα = 1/2(x -1/x)                                                      tanx = sinx/cosx

sinα × secα = 1/2(x - 1/x)

sinα [ 1/2(x + 1/x) ] = 1/2 ( x - 1/x)

sinα = (x - 1/x) / (x +1/x)

sinα = [(x² - 1)/x] / [(x²+ 1)/x]

sinα = (x² - 1 )/(x² +1);

Answer 2

Answer is non of these