If tan-sin-cos-sin-cos x- then show that sin-cos-2cos-

Tan θ= sin α cos α/sin α + cos α ⇒ tan θ = tan α 1/tan α+1 ⇒ tan θ = tan α tan π/4/1+ tan π/4.tan α⇒ tan θ= tan ( α π/4 ) ⇒θ= α π/4 ⇒ cos θ= cos ( α ...

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Standard XII

Mathematics

Trigonometric Equations

If tanθ=sinα-...

Question

If $t a n θ = \frac{s i n α - c o s α}{s i n α + c o s α},$ then show that $s i n α + c o s α = \sqrt{2} c o s θ$ .

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Solution

$t a n θ = \frac{s i n α - c o s α}{s i n α + c o s α}$
$\Rightarrow t a n θ = \frac{t a n α - 1}{t a n α + 1}$
$\Rightarrow t a n θ = \frac{t a n α - t a n \frac{π}{4}}{1 + t a n \frac{π}{4} . t a n α} \Rightarrow t a n θ = t a n (α - \frac{π}{4})$
$\Rightarrow θ = α - \frac{π}{4}$
$\Rightarrow c o s θ = c o s (α - \frac{π}{4})$
$\Rightarrow c o s θ = c o s α . c o s \frac{π}{4} + s i n α . s i n \frac{π}{4} \Rightarrow c o s θ = c o s α . \frac{1}{\sqrt{2}} + s i n α . \frac{1}{\sqrt{2}}$
$\Rightarrow c o s θ = \frac{c o s α + s i n α}{\sqrt{2}} \Rightarrow \sqrt{2} c o s θ = s i n α + c o s α$
Hence proved.

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Similar questions

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\tan θ = \frac{\sin α - \cos α}{\sin α + \cos α}

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If tanθ=sinα−cosαsinα+cosα, then show that sinα+cosα=√2cosθ.

tanθ=sinα−cosαsinα+cosα ⇒tanθ=tanα−1tanα+1 ⇒tanθ=tanα−tanπ41+tanπ4.tanα⇒tanθ=tan(α−π4) ⇒θ=α−π4 ⇒cosθ=cos(α−π4) ⇒cosθ=cosα.cosπ4+sinα.sinπ4⇒cosθ=cosα.1√2+sinα.1√2 ⇒cosθ=cosα+sinα√2⇒√2cosθ=sinα+cosα Hence proved.

If $t a n θ = \frac{s i n α - c o s α}{s i n α + c o s α},$ then show that $s i n α + c o s α = \sqrt{2} c o s θ$ .