BEMATH-If-tan-x-sec-x-b-cos-x-

Question Number 107364 by bemath last updated on 10/Aug/20

⊚ BEMATH ⊚

I f \tan x + \sec x = b \Rightarrow \cos x = ?

Answered by bobhans last updated on 10/Aug/20

⇛ BOB HANS ⇚

\frac{\tan x - \sec x}{\tan x - \sec x} \times \tan x + \sec x = ♭

\frac{\tan^{2} x - \sec^{2} x}{\tan x - \sec x} = ♭ \Leftrightarrow \frac{- 1}{\tan x - \sec x} = ♭

\tan x - \sec x = - \frac{1}{♭}

\tan x + \sec x = ♭

_______________-

- 2 \sec x = - \frac{1}{♭} - ♭ \Rightarrow \sec x = \frac{♭^{2} + 1}{2 ♭}

\Leftrightarrow ∴ \cos x = \frac{2 ♭}{♭^{2} + 1}

Answered by som(math1967) last updated on 10/Aug/20

\sec^{2} x - \tan^{2} x = 1

\Rightarrow secx - tanx = \frac{1}{b} [∵ secx + tanx = b]

∴ 2 secx = b + \frac{1}{b}

secx = \frac{b^{2} + 1}{2 b} ∴ cosx = \frac{2 b}{b^{2} + 1} ans

Answered by Dwaipayan Shikari last updated on 10/Aug/20

\sec^{2} x - \tan^{2} x = 1

b (tanx - secx) = - 1

tanx + secx = b

tanx - secx = - \frac{1}{b}

2 secx = b + \frac{1}{b}

cosx = \frac{2 b}{b^{2} + 1}

Answered by Dwaipayan Shikari last updated on 10/Aug/20

\tan^{2} x + \sec^{2} x + 2 tanxsecx = b^{2}

{2 \tan}^{2} x + 1 + 2 tanxsecx = b^{2}

2 tanx (tanx + secx) = b^{2} - 1

tanx = \frac{b^{2} - 1}{2 b}

\frac{\sin^{2} x}{\cos^{2} x} + 1 = \frac{{(b^{2} - 1)}^{2}}{{4 b}^{2}} + 1

\frac{1}{\cos^{2} x} = \frac{{(b^{2} + 1)}^{2}}{{4 b}^{2}}

cosx = \frac{2 b}{b^{2} + 1}