Question

Sin30+tan45-cosec60/sec30+cos60+cot45

Answer 1

Heya !!

I can help you. It is question from trigonometry chapter.

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We know that,

Sin30° = 1/2

Tan45° = 1

Cosec60° = 2/√3

Sec30° = 2/√3

Cos60° = 1/2

Cot45° = 1

given us,

Sin30° + Tan45° - Cosec60°/Sec30° + Cos60° + Cot45°

= 1/2 + 1 - 2/√3 / 2/√3 + 1/2 + 1

= 3/2 - 2/√3 / 2/√3 + 3/2

= 3√3 - 4 / 4 + 3√3

= 3√3 - 4 / 4 + 3√3 × 3√3 - 4 / 3√3 - 4

= 27 + 16 - 24√3 / 27 - 16

= 43 - 24√3 / 11

hopes I helped

thanks

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Answer 2

Answer: your answer is $3\sqrt{3}-4/4+3\sqrt{3}$

Explanation:

$1/2+1-2\sqrt{3}/2\sqrt{3} whole-divided-by 2\sqrt{3} +1/2+1$

$3/2-2\sqrt{3}-whole-divided-by-2/\sqrt{3}+3/2$

$\sqrt[3]{3}-4/\sqrt[2]{3}/4-whole-divided-by-4+\sqrt[3]{3}/2\sqrt{3}$$\sqrt[3]{3}-4/\sqrt[2]{3}*\sqrt[2]{3}/4+\sqrt[3]{3}$

$\sqrt[3]{3}-4/4+\sqrt[3]{3}$