Question

Man gyat karna tan30°.sec45°+tan60°.sec 30°

Answer 1

Answer:

answer for the given problem is given

Answer 2

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Q:-tan30°.sec45°+tan60°.sec 30°

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------>>>>Here is your answer<<<<--------

$⇒ \frac{1}{ \sqrt{3} } + \sqrt{3} \times \frac{2}{ \sqrt{3} }$

$⇒ \frac{ \sqrt{2} }{ \sqrt{3} } + 2$

$⇒ \frac{ \sqrt{2} + 2 \sqrt{3} }{ \sqrt{3} }$

$⇒ \frac{ \sqrt{2} + 2 \sqrt{3} }{ \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3} }$

$⇒ \frac{ \sqrt{2} \sqrt{3} + 6 }{3}$

$⇒ \frac{ \sqrt{3} ( \sqrt{2} + 2 \sqrt{3} )}{ \sqrt{3} \times \sqrt{3} }$

$⇒ \frac{ \sqrt{2} + 2 \sqrt{3} }{ \sqrt{3} } ✓$

HOPE IT HELPS YOU..

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