Question

Sin(90 plus theta)tan(270 plus theta) cos(90 plus theta) cosec(270 plus theta)

Answer 1

Answer:

sin(90 plus theta) tan (270plus theta) cos (90plus theta) cosec (270plus theta)

Explanation:

➡️Solution:

➡️Identity used:

\begin{lgathered}cos \: (90° - \theta) = sin \: \theta \\ \\ cosec \: (90° - \theta) = sec \: \theta \\ \\sin \: (90° - \theta) = cos \: \theta \\ \\ sec \: (90° - \theta) = cosec \: \theta \\ \\ {sin}^{2}\theta + {cos}^{2} \theta = 1 \\\end{lgathered}

cos(90°−θ)=sinθ

cosec(90°−θ)=secθ

sin(90°−θ)=cosθ

sec(90°−θ)=cosecθ

sin

2

θ+cos

2

θ=1

\begin{lgathered}sin \theta + cos \: \theta + sin \: \theta \: cos (90°- \theta) \frac{cos \: \theta}{sec(90° - \theta)} \\ + cos \: \theta sin (90°-\theta) \frac{sin \: \theta}{cosec(90° - \theta)} - 2 sin(90°-\theta) cos (90°-\theta) \\ \\ =sin \: \theta + cos \: \theta + sin \: \theta \: sin\theta \frac{cos \: \theta}{cosec \: \theta} + cos \: \theta \: cos\theta \: \frac{sin\theta}{sec \: \theta} - 2 \: cos\theta \: sin \: \theta \\ \\ = sin \: \theta + cos\theta + {sin}^{3} \theta \: cos\theta + {cos}^{3} \theta \: sin \: \theta - 2 \: cos\theta \: sin \: \theta \\ \\ = sin \: \theta + cos\theta + cos \: \theta \: sin \: \theta( {sin}^{2}\theta + { cos}^{2}\theta) - 2 \: cos\theta \: sin \: \theta \\ \\ = sin \: \theta + cos\theta + cos \: \theta \: sin \: \theta - 2 \: cos\theta \: sin \: \theta \\ \\ = sin \: \theta + cos\theta - cos \: \theta \: sin \: \theta \\ \\\end{lgathered}

sinθ+cosθ+sinθcos(90°−θ)

sec(90°−θ)

cosθ

+cosθsin(90°−θ)

cosec(90°−θ)

sinθ

−2sin(90°−θ)cos(90°−θ)

=sinθ+cosθ+sinθsinθ

cosecθ

cosθ

+cosθcosθ

secθ

sinθ

−2cosθsinθ

=sinθ+cosθ+sin

3

θcosθ+cos

3

θsinθ−2cosθsinθ

=sinθ+cosθ+cosθsinθ(sin

2

θ+cos

2

θ)−2cosθsinθ

=sinθ+cosθ+cosθsinθ−2cosθsinθ

=sinθ+cosθ−cosθsinθ

so,

\begin{lgathered}= sin \: \theta + cos\theta (1 - \: sin \: \theta ) \\ \\\end{lgathered}

=sinθ+cosθ(1−sinθ)

is the final solution.

➡️➡️If there is

\begin{lgathered}sin \theta cos \: \theta + sin \: \theta \: cos (90°- \theta) \frac{cos \: \theta}{sec(90° - \theta)} \\ + cos \: \theta sin (90°-\theta) \frac{sin \: \theta}{cosec(90° - \theta)} \\- 2 sin(90°-\theta) cos (90°-\theta) =0\\\end{lgathered}

sinθcosθ+sinθcos(90°−θ)

sec(90°−θ)

cosθ

+cosθsin(90°−θ)

cosec(90°−θ)

sinθ

−2sin(90°−θ)cos(90°−θ)=0

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