The problem is to simplify the following trigonometric expression:sin⁡150∘−cos⁡240∘cot⁡730∘cot⁡800∘+tan⁡730∘tan⁡800∘

Due to periodic properties of sine and cosine we have cot⁡730∘=1tan⁡730∘=cot⁡10∘=1tan⁡10∘ and cot⁡800∘=1tan⁡800∘=cot⁡80∘=1tan⁡80∘. Therefore:1cot⁡730∘cot⁡800∘+tan⁡730∘tan⁡800∘=1cot⁡10∘cot⁡80∘+tan⁡10∘tan⁡80∘Now note thatcos⁡10∘=cos⁡(−10∘)=sin⁡(−10∘+90∘)=sin⁡80∘cos⁡80∘=−sin⁡(80∘−90∘)=−sin⁡(−10∘)=sin⁡10∘sotan⁡10∘=sin⁡10∘cos⁡10∘=cos⁡80∘sin⁡80∘=cot⁡80∘,tan⁡80∘=sin⁡80∘cos⁡80∘=cos⁡10∘sin⁡10∘=cot⁡10∘,so you have1cot⁡10∘cot⁡80∘+tan⁡10∘tan⁡80∘1cot⁡10∘tan⁡10∘+tan⁡10∘cot⁡10∘=12

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Aedan Berry

Answered question

2022-04-11

The problem is to simplify the following trigonometric expression:

\frac{{\sin 150}^{\circ} - {\cos 240}^{\circ}}{{\cot 730}^{\circ} {\cot 800}^{\circ} + {\tan 730}^{\circ} {\tan 800}^{\circ}}

Answer & Explanation

akorda5cc2

Beginner2022-04-12Added 6 answers

Due to periodic properties of sine and cosine we have

{\cot 730}^{\circ} = \frac{1}{{\tan 730}^{\circ}} = {\cot 10}^{\circ} = \frac{1}{{\tan 10}^{\circ}}

and

{\cot 800}^{\circ} = \frac{1}{{\tan 800}^{\circ}} = {\cot 80}^{\circ} = \frac{1}{{\tan 80}^{\circ}}

. Therefore:

\frac{1}{{\cot 730}^{\circ} {\cot 800}^{\circ} + {\tan 730}^{\circ} {\tan 800}^{\circ}}

= \frac{1}{{\cot 10}^{\circ} {\cot 80}^{\circ} + {\tan 10}^{\circ} {\tan 80}^{\circ}}

Now note that

{\cos 10}^{\circ} = \cos (- 10^{\circ}) = \sin (- 10^{\circ} + 90^{\circ}) = {\sin 80}^{\circ}

{\cos 80}^{\circ} = - \sin (80^{\circ} - 90^{\circ}) = - \sin (- 10^{\circ}) = {\sin 10}^{\circ}

{\tan 10}^{\circ} = \frac{{\sin 10}^{\circ}}{{\cos 10}^{\circ}} = \frac{{\cos 80}^{\circ}}{{\sin 80}^{\circ}} = {\cot 80}^{\circ},

{\tan 80}^{\circ} = \frac{{\sin 80}^{\circ}}{{\cos 80}^{\circ}} = \frac{{\cos 10}^{\circ}}{{\sin 10}^{\circ}} = {\cot 10}^{\circ},

so you have

\frac{1}{{\cot 10}^{\circ} {\cot 80}^{\circ} + {\tan 10}^{\circ} {\tan 80}^{\circ}}

\frac{1}{{\cot 10}^{\circ} {\tan 10}^{\circ} + {\tan 10}^{\circ} {\cot 10}^{\circ}} = \frac{1}{2}

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