Question

Sec40° +√3 cossec40° = 4​

Answer 1

Prove that :

`\sf\sec40 \degree + √(3) \csc40 \degree = 4`

Explanation:

`\rightarrow\sf\sec40 \degree + √(3) \csc40 \degree`

`\rightarrow \: \sf (1)/( \cos 40\degree) + ( √(3) )/( \sin40 \degree )`

`\rightarrow \: \sf (\sin40 \degree + √(3)\cos 40\degree )/( \cos 40\degree. \sin40 \degree)`

Multiply and Divide ½

`\rightarrow \: \sf ( (1)/(2) \sin40 \degree + ( √(3) )/(2) \cos 40\degree )/( (1)/(2) \cos 40\degree. \sin40 \degree)`

`\rightarrow \: \sf ( ( √(3) )/(2) \cos 40\degree + (1)/(2) \sin40 \degree )/( (1)/(2) \cos 40\degree. \sin40 \degree)`

we know,

`\cos30 \degree = ( √(3) )/(2) \\ \sin30 \degree = (1)/(2) \: \: \:`

`\rightarrow \: \sf ( \cos 40\degree.\cos30 \degree + \sin40 \degree .\sin 30\degree )/( (1)/(2) \cos 40\degree. \sin40 \degree)`

• 
  `\boxed{ \bf \cos(a - b) = \cos a. \cos b + \sin a. \sin b}`

`\rightarrow \: \sf ( \cos (40\degree - 30 \degree) )/( (1)/(2) \cos 40\degree. \sin40 \degree)`

`\rightarrow \: \sf ( \cos 10 \degree )/( (1)/(2) \cos 40\degree. \sin40 \degree)`

Multiply and Divide by 4

`\rightarrow \: \sf ( 4\cos 10 \degree )/( 4 * (1)/(2) \cos 40\degree. \sin40 \degree)`

`\rightarrow \: \sf ( 4\cos 10 \degree )/( 2 \sin40 \degree.\cos 40\degree)`

• 
  `\underline{\boxed{ \rm \: \sin2 \theta = 2 \sin \theta. \cos\theta}}`

`\rightarrow \: \sf ( 4\cos 10 \degree )/( \sin ( 2 * 40 \degree))`

`\rightarrow \: \sf ( 4\cos 10 \degree )/( \sin ( 80 \degree))`

`\rightarrow \: \sf ( 4\cos 10 \degree )/( \sin ( 90 \degree- 10 \degree))`

• 
  `\underline{\boxed{\sin(90 - \theta) = \cos \theta}}`

`\rightarrow \: \sf ( 4\cos 10 \degree )/( \cos 10 \degree)`

`\rightarrow \: \sf \frac{ 4 \cancel{\cos 10 \degree} }{ \cancel{\cos 10 \degree}}`

`\rightarrow \: 4`

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