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

The correct option is A 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) ⇒θ=α π/4⇒α=θ+π/4 Hence sin α+cos α=sin(θ+π/4)+ ...

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Byju's Answer

Standard XI

Mathematics

Trigonometric Ratios of Compound Angles

If tanθ=sinα-...

Question

If $t a n θ = \frac{s i n α - c o s α}{s i n α + c o s α} t h e n s i n α + c o s α a n d s i n α - c o s α$
and must be equal to

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Solution

The correct option is A
We have $t a n θ = \frac{s i n α - c o s α}{s i n α + c o s α}$
$\Rightarrow t a n θ = \frac{s i n (α - \frac{π}{4})}{c o s (α - \frac{π}{4})} \Rightarrow t a n θ = t a n (α - \frac{π}{4})$
$\Rightarrow θ = α - \frac{π}{4} \Rightarrow α = θ + \frac{π}{4}$
Hence $s i n α + c o s α = s i n (θ + \frac{π}{4}) + c o s (θ + \frac{π}{4})$
$= \sqrt{2} c o s θ$
and $s i n α - c o s α = s i n (θ + \frac{π}{4}) - c o s (θ + \frac{π}{4})$
$= \frac{1}{\sqrt{2}} s i n θ + \frac{1}{\sqrt{2}} c o s θ - \frac{1}{\sqrt{2}} c o s θ + \frac{1}{\sqrt{2}} s i n θ$
$= \frac{2}{\sqrt{2}} s i n θ = \sqrt{2} s i n θ .$

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If tan θ=sin α−cos αsin α+cos α then sin α+cos α and sin α−cos α and must be equal to

If $t a n θ = \frac{s i n α - c o s α}{s i n α + c o s α} t h e n s i n α + c o s α a n d s i n α - c o s α$
and must be equal to