If tan - sin -cos -sin -cos - 0- then sin -cos - and sin -cos -must be equal to

We have tan θ= sin α cos α/sin α+cos α ⇒tan θ= sin(α π/4)/cos(α π/4) ⇒tan θ =tan(α π/4) Since it is given that 0 lt;θ lt;π, we can write, ⇒θ=α π/4⇒α=θ+π/4 H ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Compound Angles

If tan θ= sin...

Question

If $tan θ = \frac{sin α - cos α}{sin α + cos α}; (0 < θ < π) then sin α + cos α a n d sin α - cos α$
must be equal to

\sqrt{2} c o s θ, \sqrt{2} s i n θ

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\sqrt{2} s i n θ, \sqrt{2} c o s θ

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\sqrt{2} s i n θ, \sqrt{2} s i n θ

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\sqrt{2} c o s θ, \sqrt{2} c o s θ

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Solution

The correct option is A $\sqrt{2} c o s θ, \sqrt{2} s i n θ$
We have $tan θ = \frac{sin α - cos α}{sin α + cos α}$
$\Rightarrow tan θ = \frac{sin (α - \frac{π}{4})}{cos (α - \frac{π}{4})} \Rightarrow tan θ = tan (α - \frac{π}{4})$
Since it is given that $0 < θ < π,$ we can write,
$\Rightarrow θ = α - \frac{π}{4} \Rightarrow α = θ + \frac{π}{4}$
Hence $sin α + cos α = sin (θ + \frac{π}{4}) + cos (θ + \frac{π}{4})$
$= \frac{1}{\sqrt{2}} sin θ + \frac{1}{\sqrt{2}} cos θ + \frac{1}{\sqrt{2}} cos θ - \frac{1}{\sqrt{2}} sin θ$
$= \sqrt{2} cos θ$
and $sin α - cos α = sin (θ + \frac{π}{4}) - cos (θ + \frac{π}{4})$
$= \frac{1}{\sqrt{2}} sin θ + \frac{1}{\sqrt{2}} cos θ - \frac{1}{\sqrt{2}} cos θ + \frac{1}{\sqrt{2}} sin θ$
$= \frac{2}{\sqrt{2}} sin θ = \sqrt{2} sin θ .$

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If tan θ=sin α−cos αsin α+cos α;(0<θ<π) then sin α+cos α and sin α−cos α must be equal to

If $tan θ = \frac{sin α - cos α}{sin α + cos α}; (0 < θ < π) then sin α + cos α a n d sin α - cos α$
must be equal to