Cos-tan-22 tan-1-2 -5 sin-

LHS =cos θ (tan θ +2)(2 tan θ+1) =cos θ ( sin θ/cos θ+2 ) ( 2 sin θ/cos θ+1 ) ( ∵ tan θsin θ/cos θ ) =cos (sin θ+2 cos θ)(2 sin θ+cos θ)/cos θ. cos θ =(2 sin ...

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Byju's Answer

Standard XII

Mathematics

Basic Trigonometric Identities

cosθtanθ+22 t...

Question

$c o s θ (t a n θ + 2) (2 t a n θ + 1) = 2 s e c θ + 5 s i n θ$

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Solution

$L H S = c o s θ (t a n θ + 2) (2 t a n θ + 1)$

$= c o s θ (\frac{s i n θ}{c o s θ} + 2) (\frac{2 s i n θ}{c o s θ} + 1) (∵ t a n θ \frac{s i n θ}{c o s θ})$

$= c o s \frac{(s i n θ + 2 c o s θ) (2 s i n θ + c o s θ)}{c o s θ . c o s θ}$

$= \frac{(2 s i n^{2} θ + s i n θ c o s θ + 4 s i n θ c o s θ + 2 c o s^{2} θ)}{c o s θ}$

$= \frac{2 (s i n^{2} θ + c o s^{2} θ) + 5 s i n θ c o s θ}{c o s θ}$

$= 2 s e c θ + 5 s i n θ = R H S$ Hence proved.

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cosθ(tanθ+2)(2tanθ+1)=2secθ+5sinθ

LHS=cosθ(tanθ+2)(2tanθ+1) =cosθ(sinθcosθ+2)(2sinθcosθ+1)(∵tanθsinθcosθ) =cos(sinθ+2cosθ)(2sinθ+cosθ)cosθ.cosθ =(2sin2θ+sinθcosθ+4sinθcosθ+2cos2θ)cosθ =2(sin2θ+cos2θ)+5sinθcosθcosθ =2secθ+5sinθ=RHS Hence proved.

$c o s θ (t a n θ + 2) (2 t a n θ + 1) = 2 s e c θ + 5 s i n θ$