If tan-cos x-sin x- then the absolute value of 4-2cosx-4 is

Tan(πcos x)=(πsin x) ⇒tan(πcos x)=tan(π2 πsin x) ⇒πcos x=nπ+π2 πsin x ⇒cos x+sin x=2n+12 ,n∈ℤ ⇒1√(2)cos x+1√(2)sin x=2n+12√(2) ⇒cos(x π4)=±12√(2) (∵cos y ∈ ...

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Standard XI

Mathematics

General Solution of Trigonometric Equation

If tanπcos x=...

Question

If $tan (π cos x) = cot (π sin x),$ then the absolute value of $4 \sqrt{2} cos (x - \frac{π}{4})$ is

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Solution

$tan (π cos x) = cot (π sin x) \Rightarrow tan (π cos x) = tan (\frac{π}{2} - π sin x) \Rightarrow π cos x = n π + \frac{π}{2} - π sin x \Rightarrow cos x + sin x = \frac{2 n + 1}{2}, n \in Z \Rightarrow \frac{1}{\sqrt{2}} cos x + \frac{1}{\sqrt{2}} sin x = \frac{2 n + 1}{2 \sqrt{2}} \Rightarrow cos (x - \frac{π}{4}) = \pm \frac{1}{2 \sqrt{2}} (∵ cos y \in [- 1, 1])$
So, the absolute value of $4 \sqrt{2} cos (x - \frac{π}{4})$ is $2.$

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General Solution of Trigonometric Equation

Standard XI Mathematics

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If tan(πcosx)=cot(πsinx), then the absolute value of 4√2cos(x−π4) is

tan(πcosx)=cot(πsinx)⇒tan(πcosx)=tan(π2−πsinx)⇒πcosx=nπ+π2−πsinx⇒cosx+sinx=2n+12,n∈Z⇒1√2cosx+1√2sinx=2n+12√2⇒cos(x−π4)=±12√2 (∵cosy∈[−1,1]) So, the absolute value of 4√2cos(x−π4) is 2.

If $tan (π cos x) = cot (π sin x),$ then the absolute value of $4 \sqrt{2} cos (x - \frac{π}{4})$ is