Question

If sin 0 + cos 0 = √2 sin (90° -
0), show that cot o =√2 +1.
​

Answer 1

Given Question :-

$\tt \:  \longrightarrow \: If \: sinx \: + \: cosx \: = \sqrt{2} sin(90 - x)$

$\tt \:  \longrightarrow \: prove \: that \: cotx \: = \: \sqrt{2} \: + \: 1$

Solution:

$\tt \:  \longrightarrow \: \: sinx \: + \: cosx \: = \sqrt{2} sin(90 - x)$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \{ \because \: \bf \:  sin(90 - x) = cosx \}$

$\tt \:  \longrightarrow \: \: sinx \: + \: cosx \: = \sqrt{2} \: cosx$

$\tt \:  \longrightarrow \: \: sinx \: \: = \sqrt{2} \: cosx \: - \: cosx$

$\tt \:  \longrightarrow \: sin \: x \: = \: cosx \: ( \sqrt{2} - 1)$

$\tt \:  \longrightarrow \: \dfrac{cosx}{sinx} = \dfrac{1}{ \sqrt{2} - 1}$

$\tt \:  \longrightarrow \: cotx \: = \dfrac{1}{ \sqrt{2} - 1 } \times \dfrac{ \sqrt{2} + 1}{ \sqrt{2} + 1}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \{ \because \:\bf \:  ⟼ on \: rationalization \}$

$\tt \:  \longrightarrow \: cotx \: = \dfrac{ \sqrt{2} + 1}{ {( \sqrt{2}) }^{2} - {(1)}^{2} }$

$\tt \:  \longrightarrow \: cotx \: = \dfrac{ \sqrt{2} + 1 }{2 - 1}$

$\tt\implies \:cotx \: = \: \sqrt{2} + 1$

$\large{\boxed{\boxed{\bf{Hence, Proved}}}}$

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Additional Information:-

Relationship between sides and T ratios

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

Reciprocal Identities

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

Co-function Identities

sin (90°−x) = cos x

cos (90°−x) = sin x

tan (90°−x) = cot x

cot (90°−x) = tan x

sec (90°−x) = cosec x

cosec (90°−x) = sec x

Fundamental Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1