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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

(6)sin(180∘+A)=−sinAcos(90∘−A)=sinAtan(270∘−A)=cotAsec(540∘−A)=sec(180∘−A)=−secAcos(360∘+A)=cosAcosec(270∘+A)=−secA=(−sinA)⋅sinA⋅cotA(−secA)⋅cosA⋅(−secA)=−sinAsec2A⋅tanA⋅cotA=−sinA1cos2A=−sinA⋅cos2A(7)sin(π2−θ)=cosθcos(π2−θ)=sinθ⇒sinθ⋅cosθ⋅{cosθ⋅cosecθ+sinθ⋅secθ}=sinθ⋅cosθ⋅cosθ⋅1sinθ+sinθ⋅cosθ⋅sinθ⋅1cosθ=cos2θ+sin2θ=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

(6)−sinA.sinA.cotA−secA.cosA(−secA)=−sin2AcotAcosAsec2A=−sinAcosAsecA=−sinAcos2A

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

(7)sinθcosθ{cosθcscθ+sinθsecθ}=sinθcosθ{cosθsinθ+sinθcosθ}=cos2θ+sin2θ=1

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