Question

Prove that (sin + )2+ (cos + sec )2 = 7 +2θ + 2θ
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Answer 1

Given Statement is :-

Prove that

$\rm {(sin \theta \: + \: cosec\theta \:) }^{2} + {(cos\theta \: + sec\theta \:)}^{2} = 7 + {tan}^{2} \theta \: + {cot}^{2} \theta \:$

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$\begin{gathered}\Large{\bold{\pink{\underline{Formula \: Used \::}}}} \end{gathered}$

$\boxed{ \blue{\rm {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy}}$

$\boxed{ \blue{\rm {sin}^{2}x + {cos}^{2}x = 1 }}$

$\boxed{ \blue{\rm {sec}^{2} x = 1 + {tan}^{2}x }}$

$\boxed{ \blue{\rm {cosec}^{2}x = 1 + {cot}^{2}x }}$

$\boxed{ \blue{\rm cosecx \times sinx = 1}}$

$\boxed{ \blue{\rm secx \times cosx = 1}}$

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$\large\underline\purple{\bold{Solution :- }}$

Consider LHS,

$\rm {(sin \theta \: + \: cosec\theta \:) }^{2} + {(cos\theta \: + sec\theta \:)}^{2}$

$\rm \: = {sin}^{2} \theta \: + {cosec}^{2} \theta \: + 2sin\theta \:cosec\theta \: + {cos}^{2} \theta \: + {sec}^{2} \theta \: + 2cos\theta \:sec\theta \:$

$\rm \: = ( {sin}^{2} \theta \: + {cos}^{2} \theta \:) + 2 \times 1+ 2 \times 1+ ( {sec}^{2} \theta \: + {cosec}^{2} \theta \:)$

$\rm = 1 + 2 + 2 + 1 + {tan}^{2} \theta \: + 1 + {cot}^{2} \theta \:$

$\rm \: = 7 + {tan}^{2} \theta \: + {cot}^{2} \theta \:$

$\rm \: = RHS$

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

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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

Answer 2

Step-by-step explanation:

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