Question

sin^2 30°+cos45°-tan^2 30°​

Answer 1

$\huge{\underline{\underline{\tt{\purple{❥QUESTION}}}}}$

Q. sin² 30°+cos45°-tan²30°

$\huge{\underline{\underline{\tt{\pink{✰SOLUTION࿐}}}}}$

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✧ The value of sin30°= 1/2

✧The value of cos45° = √3/2

✧The value of tan30° = 1/√3

A/q,

➨ sin²30°+ cos45° - tan²30°

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

$= > \frac{1}{4} + \frac{ \sqrt{3} }{2} - \frac{1}{3}$

$= > \frac{3 + 6 \sqrt{3} - 4 }{12}$

$= > \frac{6 \sqrt{3} - 1 }{12}$

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Answer 2

Answer:

$\huge{\underline{\underline{\tt{\red{✧QUESTION}}}}}$

Q. sin² 30°+cos45°-tan²30°

$\huge{\underline{\underline{\tt{\blue{✮SOLUTION࿐}}}}}$

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✬ The value of sin30°= 1/2

✬The value of cos45° = √3/2

✬ The value of tan30° = 1/√3

A/q,

➨ sin²30°+ cos45° - tan²30°

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

$= > \frac{1}{4} + \frac{ \sqrt{3} }{2} - \frac{1}{3}$

$= > \frac{4 + 6 \sqrt{3} - 3}{12}$

$= > \frac{6 \sqrt{3} + 1}{12}$

✧✦✧✦✧✦✧✦✧✦✧✦✧✦✧✦✧✦✧✦✧✦✧✦✧

✰HOPE IT HELPS YOU!!࿐