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TrigonometryQuestion and Answers: Page 1

Question Number 224672 Answers: 1 Comments: 0

Question Number 224606 Answers: 1 Comments: 0

α β = 9 (( (α+β))/( (α−β))) =?

$$\:\:\: \alpha\: \beta\:=\:\mathrm{9}\:\: \\ $$$$\:\:\:\:\frac{ \left(\alpha+\beta\right)}{ \left(\alpha−\beta\right)}\:=?\: \\ $$

Question Number 224562 Answers: 0 Comments: 1

_1 =sin x , _2 =cos x _3 =tan x _n =1+cos x

$$\:\: _{\mathrm{1}} =\mathrm{sin}\:{x}\:,\: _{\mathrm{2}} =\mathrm{cos}\:{x}\: \\ $$$$ _{\mathrm{3}} =\mathrm{tan}\:{x}\: \\ $$$$ _{{n}} =\mathrm{1}+\mathrm{cos}\:{x}\: \\ $$

Question Number 224515 Answers: 1 Comments: 0

((sin 35)/(tan 56))

$$\frac{\mathrm{sin}\:\mathrm{35}}{\mathrm{tan}\:\mathrm{56}} \\ $$

Question Number 223673 Answers: 1 Comments: 0

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Question Number 222577 Answers: 1 Comments: 0

Question Number 222279 Answers: 2 Comments: 0

(a^2 −b^2 )sin θ+2abcos θ=a^2 +b^2 tan θ=??

$$\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\mathrm{sin}\:\theta+\mathrm{2}{ab}\mathrm{cos}\:\theta={a}^{\mathrm{2}} +{b}^{\mathrm{2}} \\ $$$$\mathrm{tan}\:\theta=?? \\ $$

Question Number 222025 Answers: 1 Comments: 0

(((5cos^2 (π/3)+4sec^2 (π/6)−tan^2 (π/4))/(sin^2 (π/6)+cos^2 (π/6))))=?? [easy mode]

$$\left(\frac{\mathrm{5cos}\:^{\mathrm{2}} \frac{\pi}{\mathrm{3}}+\mathrm{4sec}\:^{\mathrm{2}} \frac{\pi}{\mathrm{6}}−\mathrm{tan}\:^{\mathrm{2}} \frac{\pi}{\mathrm{4}}}{\mathrm{sin}\:^{\mathrm{2}} \frac{\pi}{\mathrm{6}}+\mathrm{cos}\:^{\mathrm{2}} \frac{\pi}{\mathrm{6}}}\right)=?? \\ $$$$\left[{easy}\:{mode}\right] \\ $$

Question Number 222022 Answers: 1 Comments: 2

If ∠P+∠Q =90^0 then prove that (√(((sin P)/(cos Q))−sin Pcos Q))=cos P

$${If}\:\angle{P}+\angle{Q}\:=\mathrm{90}^{\mathrm{0}} \:{then}\:{prove}\:{that} \\ $$$$\sqrt{\frac{\mathrm{sin}\:{P}}{\mathrm{cos}\:{Q}}−\mathrm{sin}\:{P}\mathrm{cos}\:{Q}}=\mathrm{cos}\:{P} \\ $$

Question Number 221686 Answers: 3 Comments: 2

Question Number 221578 Answers: 1 Comments: 0

(3+cos x)^2 = 4−2sin^8 x x ε [ 0, 2025π ]

$$\:\: \\ $$$$ \left(\mathrm{3}+\mathrm{cos}\:\mathrm{x}\right)^{\mathrm{2}} \:=\:\mathrm{4}−\mathrm{2sin}\:^{\mathrm{8}} \mathrm{x}\: \\ $$$$\:\:\mathrm{x}\:\epsilon\:\left[\:\mathrm{0},\:\mathrm{2025}\pi\:\right]\: \\ $$

Question Number 221411 Answers: 1 Comments: 0

Question Number 221407 Answers: 0 Comments: 0

A and B are two angles such that 0^0 <B<A<90^0 then prove geometrycaly that cos (A+B)=cos Acos B−sin Asin B

$${A}\:{and}\:{B}\:{are}\:{two}\:{angles}\:{such}\:{that}\:\mathrm{0}^{\mathrm{0}} <{B}<{A}<\mathrm{90}^{\mathrm{0}} \:{then}\:{prove}\:{geometrycaly}\:{that}\: \\ $$$$\mathrm{cos}\:\left({A}+{B}\right)=\mathrm{cos}\:{A}\mathrm{cos}\:{B}−\mathrm{sin}\:{A}\mathrm{sin}\:{B} \\ $$

Question Number 220857 Answers: 1 Comments: 0

Prove that tan 20^0 tan40^0 tan 80^0 =tan 60^0

$${Prove}\:{that}\:\mathrm{tan}\:\mathrm{20}^{\mathrm{0}} \mathrm{tan40}^{\mathrm{0}} \:\mathrm{tan}\:\mathrm{80}^{\mathrm{0}} =\mathrm{tan}\:\mathrm{60}^{\mathrm{0}} \\ $$

Question Number 220855 Answers: 1 Comments: 0

If b cos(θ+120^0 )=c cos (θ+240^0 ) then prove that b−c=−(b+c)(√3) tan θ

$${If}\:\:{b}\:\mathrm{cos}\left(\theta+\mathrm{120}^{\mathrm{0}} \right)={c}\:\mathrm{cos}\:\left(\theta+\mathrm{240}^{\mathrm{0}} \right)\:{then}\:{prove}\:{that} \\ $$$${b}−{c}=−\left({b}+{c}\right)\sqrt{\mathrm{3}}\:\mathrm{tan}\:\theta \\ $$

Question Number 220745 Answers: 3 Comments: 0

Question Number 220712 Answers: 0 Comments: 0

prove: ((2 tan 2A + tan A)/(4 tan 3A − tan 2A)) = sin^2 A

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{prove}: \\ $$$$\:\:\frac{\mathrm{2}\:{tan}\:\mathrm{2}{A}\:+\:{tan}\:{A}}{\mathrm{4}\:{tan}\:\mathrm{3}{A}\:−\:{tan}\:\mathrm{2}{A}}\:=\:{sin}^{\mathrm{2}} \:{A} \\ $$$$\: \\ $$

Question Number 220560 Answers: 3 Comments: 0

sinα=0.8 ⇒ ((BE)/(EF))=?

$${sin}\alpha=\mathrm{0}.\mathrm{8}\:\Rightarrow\:\frac{{BE}}{{EF}}=? \\ $$

Question Number 220511 Answers: 3 Comments: 0

AB=2CE & DE=2(√3)+4 CE ⊥AB & AD⊥BC & AB=AC & EF ⊥BC BF=?

$${AB}=\mathrm{2}{CE}\:\:\&\:\:{DE}=\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{4}\:\:\: \\ $$$${CE}\:\bot{AB}\:\:\:\&\:\:\:{AD}\bot{BC}\:\:\&\:\:{AB}={AC}\:\:\&\:{EF}\:\bot{BC}\: \\ $$$${BF}=? \\ $$$$ \\ $$

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Question Number 217783 Answers: 0 Comments: 11

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Question Number 217384 Answers: 1 Comments: 0

Prove that sin351° = − (√(((4 − (√(10 + 2(√5))))/8) .))

$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{sin351}°\:=\:−\:\sqrt{\frac{\mathrm{4}\:−\:\sqrt{\mathrm{10}\:+\:\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{8}}\:.} \\ $$

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