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Home Question Bank Math Trigonometry Archive 2025 January 14

Trigonometry Questions from Jan 14,2025

Browse the Trigonometry Q&A Archive for Jan 14,2025, featuring a collection of homework questions and answers from this day. Find detailed solutions to enhance your understanding.

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simplify \( \frac{\sin 210 \cos 300 \tan 240}{\cos 120 \tan 150 \sin 330} \) \( 6.2 \quad[\sin (-\theta)+\cos (360-\theta)]\left[\cos (90-\theta)+\frac{\sin \theta}{\tan \theta}\right] \) Simplify the following expressions without using a calculator. \( 6.1 \quad \frac{\sin 210^{\circ} \cos 300^{\circ} \tan 240^{\circ}}{\cos 120^{\circ} \tan 150^{\circ} \sin 330^{\circ}} \) 6.1 \( \frac{\sin 210^{\circ} \cos 300^{\circ} \tan 240^{\circ}}{\cos 120^{\circ} \tan 150^{\circ} \sin 330^{\circ}} \) \( 6.2[\sin (-\theta)+\cos (360-\theta)]\left[\cos (90-\theta)+\frac{\sin \theta}{\tan \theta}\right] \) 6.3 If \( \tan x=m+\frac{1}{m}, 90^{\circ} \leq x \leq 270^{\circ} \) and \( m^{2}+\frac{1}{m^{2}}=1 \) Calculate the value of \( x \) without the use of a calculator 6. At a point 100 m from the base of the Eiffel tower, the angle of elevation of the top of the tower is \( 73^{\circ} \). How tall is the tower to the nearest metre? Simplify the following expressions without using a calculator. \( \frac{\sin 210^{\circ} \cos 300^{\circ} \tan 240^{\circ}}{\cos 120^{\circ} \tan 150^{\circ} \sin 330^{\circ}} \) e) \( \operatorname{cotg}(2 x)=4 \) Show that, \( \frac{1+\cos \theta}{1-\cos \theta}+\frac{1-\cos \theta}{1+\cos \theta}=4 \cot ^{2} \theta+2 \) 3. A road rises 15 m for each 150 m of horizontal distance. What is the angle of inclination of the road to the nearest degree? i) Simplify the following to one trigonometric ratio: i. \( \frac{\cos \left(360^{\circ}-x\right) \cdot \sin \left(180^{\circ}-x\right)}{\cos \left(180^{\circ}+x\right)} \) ii. \( \frac{\cos \left(720^{\circ}-x\right) \cdot \cos \left(90^{\circ}-x\right)}{\cos \left(-180^{\circ}-x\right)} \) iii. \( \frac{\cos \left(360^{\circ}+x\right) \cdot \cos \left(90^{\circ}-x\right)}{\cos \left(180^{\circ}-x\right) \cdot \cos \left(-180^{\circ}+x\right)} \) iv. \( \frac{\cos ^{2}\left(90^{\circ}+x\right)}{\tan \left(540^{\circ}+x\right) \cdot \cos \left(-180^{\circ}+x\right) \cdot \cos \left(180^{\circ}-x\right)} \)
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