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Question-94071

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Question Number 94071 by oustmuchiya@gmail.com last updated on 16/May/20
Commented by hknkrc46 last updated on 16/May/20
(6) sin (180^○ +A)=−sin A cos (90^○ −A)=sin A tan (270^○ −A)=cot A sec (540^○ −A)=sec (180^○ −A)=−sec A cos (360^○ +A)=cos A cosec (270^○ +A)=−sec A =(((−sin A)∙sin A∙cot A)/((−sec A)∙cos A∙(−sec A)))=((−sin A)/(sec^2 A))∙tan A∙cot A =((−sin A)/(1/(cos^2 A)))=−sin A∙cos^2 A (7) sin ((π/2)−θ)=cos θ cos ((π/2)−θ)=sin θ ⇒ sin θ∙cos θ∙{cos θ∙cosec θ+sin θ∙sec θ} =sin θ∙cos θ∙cos θ∙(1/(sin θ))+sin θ∙cos θ∙sin θ∙(1/(cos θ)) =cos^2 θ+sin^2 θ=1
$$\left(\mathrm{6}\right)\:\mathrm{sin}\:\left(\mathrm{180}^{\circ} +\mathrm{A}\right)=−\mathrm{sin}\:\mathrm{A} \\ $$$$\mathrm{cos}\:\left(\mathrm{90}^{\circ} −\mathrm{A}\right)=\mathrm{sin}\:\mathrm{A} \\ $$$$\mathrm{tan}\:\left(\mathrm{270}^{\circ} −\mathrm{A}\right)=\mathrm{cot}\:\mathrm{A} \\ $$$$\mathrm{sec}\:\left(\mathrm{540}^{\circ} −\mathrm{A}\right)=\mathrm{sec}\:\left(\mathrm{180}^{\circ} −\mathrm{A}\right)=−\mathrm{sec}\:\mathrm{A} \\ $$$$\mathrm{cos}\:\left(\mathrm{360}^{\circ} +\mathrm{A}\right)=\mathrm{cos}\:\mathrm{A} \\ $$$$\mathrm{cosec}\:\left(\mathrm{270}^{\circ} +\mathrm{A}\right)=−\mathrm{sec}\:\mathrm{A} \\ $$$$=\frac{\left(−\mathrm{sin}\:\mathrm{A}\right)\centerdot\mathrm{sin}\:\mathrm{A}\centerdot\mathrm{cot}\:\mathrm{A}}{\left(−\mathrm{sec}\:\mathrm{A}\right)\centerdot\mathrm{cos}\:\mathrm{A}\centerdot\left(−\mathrm{sec}\:\mathrm{A}\right)}=\frac{−\mathrm{sin}\:\mathrm{A}}{\mathrm{sec}\:^{\mathrm{2}} \mathrm{A}}\centerdot\mathrm{tan}\:\mathrm{A}\centerdot\mathrm{cot}\:\mathrm{A} \\ $$$$=\frac{−\mathrm{sin}\:\mathrm{A}}{\frac{\mathrm{1}}{\mathrm{cos}\:^{\mathrm{2}} \mathrm{A}}}=−\mathrm{sin}\:\mathrm{A}\centerdot\mathrm{cos}\:^{\mathrm{2}} \mathrm{A} \\ $$$$\left(\mathrm{7}\right)\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\theta\right)=\mathrm{cos}\:\theta \\ $$$$\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}−\theta\right)=\mathrm{sin}\:\theta \\ $$$$\Rightarrow\:\mathrm{sin}\:\theta\centerdot\mathrm{cos}\:\theta\centerdot\left\{\mathrm{cos}\:\theta\centerdot\mathrm{cosec}\:\theta+\mathrm{sin}\:\theta\centerdot\mathrm{sec}\:\theta\right\} \\ $$$$=\mathrm{sin}\:\theta\centerdot\mathrm{cos}\:\theta\centerdot\mathrm{cos}\:\theta\centerdot\frac{\mathrm{1}}{\mathrm{sin}\:\theta}+\mathrm{sin}\:\theta\centerdot\mathrm{cos}\:\theta\centerdot\mathrm{sin}\:\theta\centerdot\frac{\mathrm{1}}{\mathrm{cos}\:\theta} \\ $$$$=\mathrm{cos}\:^{\mathrm{2}} \theta+\mathrm{sin}\:^{\mathrm{2}} \theta=\mathrm{1} \\ $$
Commented by i jagooll last updated on 16/May/20
(6) ((−sin A.sin A.cot A)/(−sec A.cos A(−sec A))) = −((sin^2 A cot A )/(cos A sec^2 A)) =− ((sin Acos A)/(sec A)) = − sin A cos^2 A
$$\left(\mathrm{6}\right)\:\frac{−\mathrm{sin}\:\mathrm{A}.\mathrm{sin}\:\mathrm{A}.\mathrm{cot}\:\mathrm{A}}{−\mathrm{sec}\:\:\mathrm{A}.\mathrm{cos}\:\mathrm{A}\left(−\mathrm{sec}\:\mathrm{A}\right)}\:= \\ $$$$−\frac{\mathrm{sin}^{\mathrm{2}} \mathrm{A}\:\mathrm{cot}\:\mathrm{A}\:}{\mathrm{cos}\:\mathrm{A}\:\mathrm{sec}\:^{\mathrm{2}} \mathrm{A}}\:=−\:\frac{\mathrm{sin}\:\mathrm{Acos}\:\mathrm{A}}{\mathrm{sec}\:\mathrm{A}} \\ $$$$=\:−\:\mathrm{sin}\:\mathrm{A}\:\mathrm{cos}\:^{\mathrm{2}} \mathrm{A} \\ $$
Answered by i jagooll last updated on 16/May/20
(7) sin θcos θ{cos θ csc θ + sin θsec θ}= sin θcos θ{((cos θ)/(sin θ))+((sin θ)/(cos θ))} = cos^2 θ + sin^2 θ = 1
$$\left(\mathrm{7}\right)\:\mathrm{sin}\:\theta\mathrm{cos}\:\theta\left\{\mathrm{cos}\:\theta\:\mathrm{csc}\:\theta\:+\:\mathrm{sin}\:\theta\mathrm{sec}\:\theta\right\}= \\ $$$$\mathrm{sin}\:\theta\mathrm{cos}\:\theta\left\{\frac{\mathrm{cos}\:\theta}{\mathrm{sin}\:\theta}+\frac{\mathrm{sin}\:\theta}{\mathrm{cos}\:\theta}\right\}\:= \\ $$$$\mathrm{cos}\:^{\mathrm{2}} \theta\:+\:\mathrm{sin}\:^{\mathrm{2}} \:\theta\:=\:\mathrm{1} \\ $$

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