If t a n θ = s i n α − c o s α s i n α + c o s α , then s i n α + c o s α and s i n α − c o s α must be equal to Options: A . &nbsp √ 2 c o s θ , √ 2 s i n θ B . &nbsp √ 2 s i n θ , √ 2 c o s θ C . &nbsp √ 2 s i n θ , √ 2 s i n θ D . &nbsp √ 2 c o s θ , √ 2 c o s θ

Answer: Option A : A We have t a n θ = s i n α − c o s α s i n α + c o s α ⇒ t a n θ = s i n ( α − π 4 ) c o s ( α − π 4 ) ⇒ t a n θ = t a n ( α − π 4 ) ⇒ θ = α - π 4 ⇒ α = θ + π 4 Hence, s i n α + c o s α = s i n ( θ + π 4 ) + c o s ( θ + π 4 ) = √ 2 c o s θ And s i n α − c o s α = s i n ( θ + π 4 ) - c o s ( θ + π 4 ) = 1 √ 2 s i n θ + 1 √ 2 c o s θ - 1 √ 2 c o s θ + 1 √ 2 s i n θ = 2 √ 2 s i n θ = √ 2 s i n θ = √ 2 s i n θ .

If tanθ = sinα−cosαsinα+cosα, then

Discussion Forum : Trigonometric Functions

Question -

If $t a n θ$ = $\frac{s i n α - c o s α}{s i n α + c o s α}$ , then $s i n α + c o s α$ and

$s i n α - c o s α$ must be equal to

Options:

A .

\sqrt{2} c o s θ, \sqrt{2} s i n θ

B .

\sqrt{2} s i n θ, \sqrt{2} c o s θ

C .

\sqrt{2} s i n θ, \sqrt{2} s i n θ

D .

\sqrt{2} c o s θ, \sqrt{2} c o s θ

Answer: Option A
:
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 $s i n α - c o s α$ = $s i n (θ + \frac{π}{4})$ - $c o s (θ + \frac{π}{4})$

= $\frac{1}{\sqrt{2}}$ $s i n θ$ + $\frac{1}{\sqrt{2}}$ $c o s θ$ - $\frac{1}{\sqrt{2}}$ $c o s θ$ + $\frac{1}{\sqrt{2}}$ $s i n θ$

= $\frac{2}{\sqrt{2}}$ $s i n θ$ = $\sqrt{2} s i n θ = \sqrt{2} s i n θ$ .

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Discussion Forum : Trigonometric Functions

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