Evaluate :cos40∘sin50∘+sec50∘cosec40∘−4cos50∘cosec40∘

We know,cos(90−θ)=sinθsec(90−θ)=cscθcosecθ=sinθ1cos40∘=cos(90∘−50∘)=sin50∘sec50∘=tan(90∘−40∘)=csc40∘cos50∘=sin(90∘−40∘)=sin40∘To evaluate,cos40∘sin50∘+sec50∘csc40∘−4cos50∘csc40∘=sin50∘sin50∘+csc40∘csc40∘−4(sin40∘cos50∘)=1+1−4(sin40∘sin40∘)=1+1−4=−2

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Evaluate : { , 50^{}}{ , 40^{}} , + , {cosec ,40^{}}{ , 50^{}}

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Evaluate :
$\frac{sin 5 0 ^{\circ}}{cos 4 0 ^{\circ}} + \frac{cosec 4 0 ^{\circ}}{sec 5 0 ^{\circ}} - 4 cos 5 0^{\circ} cosec 4 0^{\circ}$

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Evaluate :

\frac{sin 5 0 ^{\circ}}{cos 4 0 ^{\circ}} + \frac{cosec 4 0 ^{\circ}}{sec 5 0 ^{\circ}} - 4 cos 5 0^{\circ} cosec 4 0^{\circ}

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Text solutionVerified

We know,

cos (90 - θ) = sin θ

sec (90 - θ) = csc θ

cosec θ = \frac{1}{sin θ}

cos 4 0^{\circ} = cos (9 0^{\circ} - 5 0^{\circ}) = sin 5 0^{\circ}

sec 5 0^{\circ} = tan (9 0^{\circ} - 4 0^{\circ}) = csc 4 0^{\circ}

cos 5 0^{\circ} = sin (9 0^{\circ} - 4 0^{\circ}) = sin 4 0^{\circ}

To evaluate,

\frac{sin 5 0 ^{\circ}}{cos 4 0 ^{\circ}} + \frac{csc 4 0 ^{\circ}}{sec 5 0 ^{\circ}} - 4 cos 5 0^{\circ} csc 4 0^{\circ}

= \frac{sin 5 0 ^{\circ}}{sin 5 0 ^{\circ}} + \frac{csc 4 0 ^{\circ}}{csc 4 0 ^{\circ}} - 4 (\frac{cos 5 0 ^{\circ}}{sin 4 0 ^{\circ}})

= 1 + 1 - 4 (\frac{sin 4 0 ^{\circ}}{sin 4 0 ^{\circ}})

= 1 + 1 - 4

= - 2

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Question Text	Evaluate : $\frac{sin 5 0 ^{\circ}}{cos 4 0 ^{\circ}} + \frac{cosec 4 0 ^{\circ}}{sec 5 0 ^{\circ}} - 4 cos 5 0^{\circ} cosec 4 0^{\circ}$
Answer Type	Text solution:1
Upvotes	150

Evaluate :cos40∘sin50∘​+sec50∘cosec40∘​−4cos50∘cosec40∘

Text solutionVerified

Practice questions from similar books

Evaluate :
$\frac{sin 5 0 ^{\circ}}{cos 4 0 ^{\circ}} + \frac{cosec 4 0 ^{\circ}}{sec 5 0 ^{\circ}} - 4 cos 5 0^{\circ} cosec 4 0^{\circ}$