Question

If sec θ + tan θ = x, then sec θ =
(a)$\frac{x^{2}+1} {x}$
(b)$\frac{x^{2}+1} {2x}$
(c)$\frac{x^{2}-1} {2x}$
(d)$\frac{x^{2}-1} {x}$

Answer 1

Answer:

The value of sec θ is (x² + 1)/2x.

Among the given options option (b) (x² + 1)/2x is correct.  

Step-by-step explanation:

Given : sec θ + tan θ = x …….. (1)

By using an identity , sec² θ - tan² θ = 1

(sec θ + tan θ)(sec θ -  tan θ) = 1

[By using identity , a² - b² = (a + b) (a - b) ]

x (sec θ -  tan θ) = 1

(sec θ -  tan θ) = 1/x ……….(2)

On adding eq 1 & 2,  

sec θ + tan θ + sec θ -  tan θ = (x + 1/x)

2 sec θ = (x + 1/x)

2 sec θ = (x² + 1)/x

sec θ = (x² + 1)/2x

Hence, the value of sec θ is (x² + 1)/2x.

HOPE THIS ANSWER WILL HELP YOU…

Answer 2

Answer:

Option (c)

Step-by-step explanation:

Given : secθ + tanθ = x ...(i)

{ As we know that,

sec²θ - tan²θ = 1 }

Now, (secθ - tanθ)(secθ + tanθ) = 1

[ a² - b² = (a - b)(a + b)

[ a² - b² = (a - b)(a + b) Here, a = secθ, b = tanθ ]

(secθ - tanθ)(x) = 1 [using (i)]

secθ - tanθ = 1/x ...(ii)

Subtracting (i) and (ii), we get

→ (secθ + tanθ) - (secθ - tanθ) = x - 1/x

→ secθ + tanθ - secθ + tanθ = (x² - 1)/x

→ 2 tanθ = (x² - 1)/x

tan θ = (x² - 1)/2x