Question

3 sin alpha-4cos alpha=0 what is sec alpha

Answer 1

$\star\:\:\:\sf\large\underline\blue{Question:-}$

• If $\sf3 \sin \alpha - 4 \cos \alpha = 0$, find the value of $\sf \sec \alpha$

$\star\:\:\:\sf\large\underline\blue{Solution:-}$

Given,

$\sf3 \sin \alpha - 4 \cos \alpha = 0$

$\sf \implies 3 \sin \alpha = 4 \cos \alpha$

$\sf \implies \tan \alpha = \frac{4}{3}$

Now, we know that,

$\sf{ \sec }^{2} \alpha - { \tan }^{2} \alpha = 1$

$\sf \implies{ \sec }^{2} \alpha = 1 + { \tan }^{2} \alpha$

$\sf \implies{ \sec } \: \alpha = \sqrt{ 1 + { \tan }^{2} \alpha }$

$\sf \sqrt{1 + \frac{16}{9} }$

$\sf= \sqrt{ \frac{9 + 16}{9} }$

$\sf= \sqrt{ \frac{25}{9} }$

$\sf = \frac{5}{3}$

$\star\:\:\:\sf\large\underline\blue{Answer:-}$

• $\sf \sec \: \alpha = \frac{5}{3}$

Answer 2

Answer:

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