Question

sec thita +tan titha = 3/2 value of cos titha ?

Answer 1

Answer:

$\large{\underline{\boxed{\sf \cos \theta = \dfrac{12}{13}}}}$

Explanation:

Given that ;

$\sec \theta + \tan \theta = \dfrac{3}{2} \: \: \: .....(i)$

We know that,

$\sec^{2} \theta - { \tan}^{2} \theta = 1$

Also,

• a² - b² = (a + b)(a - b)

${ \sec}^{2}\theta - { \tan}^{2} \theta = 1 \\ \\ \rightarrow ( \sec \theta + \tan \theta) ( \sec \theta - \tan \theta) = 1 \\ \\ \rightarrow \frac{3}{2} \times ( \sec \theta - \tan \theta) = 1 \\ \\ \rightarrow ( \sec \theta - \tan \theta) = \frac{2}{3} \: \: \: .....(ii)$

On adding equation (i) and (ii),

$\sec \theta + \tan \theta + \sec \theta - \tan \theta = \frac{3}{2} + \frac{2}{3} \\ \\ 2 \sec \theta = \frac{9 + 4}{6} \\ \\ \sec \theta = \frac{13}{6} \times \frac{1}{2} \\ \\ \sec \theta = \frac{13}{12}$

$\rule{300}{2}$

We know that,

Sec $\theta$ = $\rm{\dfrac{Hypotenuse}{Base}}$

We get,

• Hypotenuse = 13
• Base = 12

We need to find, perpendicular -

$\rm perpendicular = \sqrt{h {}^{2} - b {}^{2} } \\ \\ \implies \rm p = \sqrt{(13) {}^{2} - (12) {}^{2} } \\ \\ \implies \rm p = \sqrt{169 - 144} \\ \\ \implies \rm p = \sqrt{25} \\ \\ \implies \rm p = 5$

Thus, the perpendicular is 5.

$\cos \theta = \rm \frac{base}{hypotenuse} \\ \\ \implies \cos \theta = \frac{12}{13}$

Hence, the answer is 12/13.