Question

if cot theta = 1 then evaluate 1+cos theta / sin theta​

Answer 1

$Given \: cot \:\theta = 1$

$\implies cot \:\theta = cot \:45\degree$

$\implies \theta = 45\degree\: ---(1)$

$Value \: of \: \frac{ 1+ cos\:\theta }{sin \:\theta}$

$= \frac{ 1+ cos\:45\degree }{sin \:45\degree}$

$=\frac{ 1 + \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}$

$= \frac{\frac{\sqrt{2}+1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}$

$= \sqrt{2} + 1 \\= 1.414 + 1 \\= 2.414$

Therefore.,

$\red {Value \: of \: \frac{ 1+ cos\:\theta }{sin \:\theta}}\green {= 2.414 }$

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Answer 2

Answer:

Given, cot θ = 7/8

=> tan θ = 1/(7/8) {since tan θ = 1/cot θ}

=> tan θ = 8/7

In the figure,

from triangle ABC,

Apply Pythagoras Theorem, we get

=> (Hypotanous)2 = (Perpendicular)2 + (Base)2

=> AC2 = AB2 + BC2

=> AC2 = 82 + 72

=> AC2 = 64 + 49

=> AC2 = 113

=> AC = √113

Given, {(1 + sin θ)*(1 - sin θ)}/{(1 + cos θ)*(1 - cos θ)}

= (1 - sin2 θ)}/{(1 - cos2 θ) {since a2 - b2 = (a - b)*(a + b)}

= (1 - sin2 θ)}/sin2 θ {since sin2 θ + cos2 θ = 1 }

Grom the figure, sin θ = AB/AC = 8/√113

Now, (1 - sin2 θ)}/sin2 θ = {1 - (8/√113)2 )}/(8/√113)2

= (1 - 64/113)/(64/113)

= {(113 - 64)/113}/(64/113)

= (49/113)/(64/113)

= 49/64