Question

If secθ+tanθ=31, then tanθ-secθ value is how much ?

Answer 1

Find the value:

Step-by-step explanation:

$\ Given \ value: \\\\\sec\theta + \tan\theta = 31 \\\\\ find: \\\\\sec\theta - \tan\theta = ?\\\\\ Solution: \\\\\sec\theta + \tan\theta = 31 \\\\\rightarrow \sec\theta \ = 31 - \tan\theta\\\\\ Square \ the \ above \ value: \\\\\rightarrow (\sec\theta)^2 \ = (31 - \tan\theta)^2\\\\\rightarrow \sec^2 \theta \ = (31)^2 + \tan^2\theta -2\cdot31\tan\theta\\\\\rightarrow \ 1\ + tan^2 \theta \ = (31)^2 + \tan^2\theta -2\cdot31\tan\theta$

$\rightarrow \ 1 \ = (31)^2 -2\cdot31\tan\theta\\\\\rightarrow \ 1 \ = 961 - 62\tan\theta\\\\\rightarrow \ 1 \ - 961 = - 62\tan\theta\\\\\rightarrow - 960 = - 62\tan\theta\\\\\rightarrow - 62\tan\theta= - 960 \\\\\rightarrow \tan\theta\ = \frac{960}{62}$

$\ tan\theta = \frac{p}{b} \\\\ \sec\theta= \frac{h}{b} \\\\h^2=p^2+b^2 \\\\h^2= (960)^2+(62)^2 \\\\h^2= 921600+3,844\\\\h^2= 921600+3,844\\\\h^2=925444\\\\h^2=(962)^2\\\\ h=962$

$\ sec \ \theta = \frac {962}{62} \\\\\rightarrow \tan\theta- \sec\theta \\\\\ put \ the \ value \ of \ sec \theta \ and \ \tan \theta \ value \ in \ above \ equation \\\\ \rightarrow \frac {960}{62} - \frac {962}{62} \\\\\rightarrow \frac {960-962}{62} \\\\\rightarrow - \frac {2}{62} \\\\\rightarrow - \frac {1}{31}$

$\tan \theta - \ sec \ \theta = - \frac {1}{31}$

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• Proving: https://brainly.in/question/8112701