Question

3 tan^2 30° + sec^2 45° - tan^2 60° is equal to:

Answer 1

${\red{\boxed{\boxed{\huge{\bold{Solution:-}}}}}}$

We know that,

$\tan(30) = \frac{1}{ \sqrt{3} } \\ \sec(45) = \sqrt{2} \\ \tan(60) = \sqrt{3}$

3tan²30°+sec²45°-tan²60°

$= 3 \times { (\frac{1}{ \sqrt{3} } )}^{2} + {( \sqrt{2}) }^{2} - { (\sqrt{3} )}^{2} \\ = 1 + 2 - 3 \\ = 3 - 3 \\ = 0$

${\red{\boxed{\boxed{\huge{\bold{Answer:-}}}}}}$

Value of 3 tan² 30° + sec² 45° - tan² 60° is 0.