If a tan θ = b, then a cos 2 θ + b sin 2 θ = Options: A . &nbsp a B . &nbsp b C . &nbsp -a D . &nbsp -b

Answer: Option A : A Given that tan θ = b a . Now, a cos 2 θ + b sin 2 θ = a ( 1 − t a n 2 θ 1 + t a n 2 θ ) + b ( 2 t a n θ 1 + t a n 2 θ ) Putting t a n θ = b a , we get = a ⎛ ⎝ 1 − b 2 a 2 1 + b 2 a 2 ⎞ ⎠ + b ⎛ ⎝ 2 b a 1 + b 2 a 2 ⎞ ⎠ = a ( a 2 − b 2 a 2 + b 2 ) + b ( 2 b a a 2 + b 2 ) = 1 ( a 2 + b 2 ) a 3 − a b 2 + 2 a b 2 = a ( a 2 + b 2 ) a 2 + b 2 = a.

If a tan θ = b, then a cos 2θ + b sin 2θ =

Discussion Forum : Trigonometric Functions

Question -

If a tan $θ$ = b, then a cos 2 $θ$ + b sin 2 $θ$ =

Options:

A . a

B . b

C . -a

D . -b

Answer: Option A
:
A

Given that tan $θ$ = $\frac{b}{a}$ .

Now, a cos 2 $θ$ + b sin 2 $θ$ = a $(\frac{1 - t a n^{2} θ}{1 + t a n^{2} θ})$ + b $(\frac{2 t a n θ}{1 + t a n^{2} θ})$

Putting $t a n θ$ = $\frac{b}{a}$ , we get

= a $⎛ ⎝ \frac{1 - \frac{b^{2}}{a^{2}}}{1 + \frac{b^{2}}{a^{2}}} ⎞ ⎠$ + b $⎛ ⎝ \frac{2 \frac{b}{a}}{1 + \frac{b^{2}}{a^{2}}} ⎞ ⎠$ = a $(\frac{a^{2} - b^{2}}{a^{2} + b^{2}})$ + b $(\frac{2 b a}{a^{2} + b^{2}})$

= $\frac{1}{(a^{2} + b^{2})}$ $a^{3} - a b^{2} + 2 a b^{2}$ = $\frac{a (a^{2} + b^{2})}{a^{2} + b^{2}}$ = a.

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