Question

$\bold{if \: {cos}^{2 } \theta + 2 {sin}^{2} \theta + 3 {cos}^{2} \theta + {4sin}^{2} \theta } \\ \bold{... \: upto \: 200 \: terms = 10025 \: where} \\ \bold{ \theta \: is \: acute. \: then\: find \: (sin \theta - cos \theta) }$
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Answer 1

$\large\underline{\sf{Solution-}}$

Given that

$\rm \: {cos}^{2}\theta+2{sin}^{2}\theta + {3cos}^{2}\theta + {4sin}^{2}\theta + - - 200 \: terms = 10025$

On rearranging the terms, we get

$\rm \:({cos}^{2}\theta + {3cos}^{2}\theta + - - - 100 \: terms) + ( {2sin}^{2}\theta + {4sin}^{2}\theta + - - - 100 \: terms) = 10025$

can be further rewritten as

$\rm \: {cos}^{2}\theta(1 + 3 + 5 +- - 100 \: terms) + {sin}^{2}\theta(2 + 4 + 6 + - - 100 \: terms) = 10025$

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ Sum of n  terms of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{S_n\:=\dfrac{n}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg)}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

• Sₙ is the sum of n terms of AP.

• a is the first term of the sequence.

• n is the no. of terms.

• d is the common difference.

So, using this formula we get

$\rm \: {cos}^{2}\theta \times \dfrac{100}{2} \bigg(2(1) + (100 - 1)2 \bigg) + {sin}^{2}\theta \times \dfrac{100}{2} \bigg(2(2) + (100 - 1)2 \bigg) = 10025$

$\rm \: 50{cos}^{2}\theta \bigg(2 + 198 \bigg) + {50sin}^{2}\theta\bigg(4 + 198 \bigg) = 10025$

$\rm \: 50{cos}^{2}\theta \bigg(200 \bigg) + {50sin}^{2}\theta\bigg(200 + 2 \bigg) = 10025$

$\rm \: 10000{cos}^{2}\theta + 10000{sin}^{2}\theta + 100 {sin}^{2}\theta = 10025$

$\rm \: 10000({cos}^{2}\theta + {sin}^{2}\theta) + 100 {sin}^{2}\theta = 10025$

$\rm \: 10000(1) + 100 {sin}^{2}\theta = 10025$

$\rm \: 10000 + 100 {sin}^{2}\theta = 10025$

$\rm \: 100 {sin}^{2}\theta = 10025 - 10000$

$\rm \: 100 {sin}^{2}\theta = 25$

$\rm \:{sin}^{2}\theta = \dfrac{25}{100}$

$\rm \:{sin}^{2}\theta = \dfrac{1}{4}$

$\rm \implies\:sin\theta = \dfrac{1}{2}$

$\bf\implies \:\theta = 30 \degree \:$

Now, Consider

$\red{\rm :\longmapsto\:sin\theta - cos\theta \: }$

$\rm \:  =  \: sin30 \degree - cos30 \degree$

$\rm \:  =  \: \dfrac{1}{2} - \dfrac{ \sqrt{3} }{2}$

$\rm \:  =  \: \dfrac{1 - \sqrt{3} }{2}$

Hence,

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: sin\theta - cos\theta \: } = \frac{1 - \sqrt{3} }{2} \: }}$

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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

Given that

On rearranging the terms, we get

can be further rewritten as

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ Sum of n  terms of an arithmetic sequence is,

Wʜᴇʀᴇ,

Sₙ is the sum of n terms of AP.

a is the first term of the sequence.

n is the no. of terms.

d is the common difference.

So, using this formula we get

Now, Consider

Hence,

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

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