Question

If sin a=1/2 find the value of 2sec a/1+tansqure a

Answer 1

Answer:√3

Step-by-step explanation:

sin a=1/2 implies a may be 30°.

Now, 2sec a/(1+tan^2 a)

can be written as 2sec a/ sec^2a

(Since 1+tan^2 a is sec^2 a)

Simplyfiying it turns to 2/sec a which is 2cos a (since 1/sec a= cos a)

Substitution of a as 30° gives 2x(√3/2) =√3.

Answer 2

Answer:

$Value \:of \: \frac{2secA}{1+tan^{2}A}=\sqrt{3}$

Step-by-step explanation:

$Given \: sinA= \frac{1}{2}$

$\implies sinA= sin 30\degree$

$\implies A = 30\degree\:--(1)$

$Now,\\\frac{2secA}{1+tan^{2}A}\\=\frac{2secA}{sec^{2}A}$

/* By trigonometric identity:

1+tan²A = sec²A */

$= \frac{2}{secA}\\=2cosA\\=2cos30\degree\\=2\times \frac{\sqrt{3}}{2}\\=\sqrt{3}$

Therefore,

$Value \:of \: \frac{2secA}{1+tan^{2}A}=\sqrt{3}$

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