Question

EVALUATE : SIN THETA + COS THETA + SIN THETA COS (90°- THETA) COS THETA / SEC(90°-theta) + cos theta sin (90-theta)sin theta /cosec(90°-theta) - 2 sin(90°-theta) cos (90°-theta)

Answer 1

➡️Solution:

➡️Identity used:

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

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

so,
$= sin \: \theta + cos\theta (1 - \: sin \: \theta )$
is the final solution.

➡️➡️If there is
$sin \theta cos \: \theta + sin \: \theta \: cos (90°- \theta) \frac{cos \: \theta}{sec(90° - \theta)} \\ + cos \: \theta sin (90°-\theta) \frac{sin \: \theta}{cosec(90° - \theta)} \\- 2 sin(90°-\theta) cos (90°-\theta) =0$

Answer 2

Answer:

cos(90°−θ)=sinθ

cosec(90°−θ)=secθ

sin(90°−θ)=cosθ

sec(90°−θ)=cosecθ

sin

2

θ+cos

2

θ=1

\begin{gathered}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{gathered}

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

=sinθ+cosθ(1−sinθ)

is the final solution.

➡️➡️If there is

\begin{gathered}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{gathered}

sinθcosθ+sinθcos(90°−θ)

sec(90°−θ)

cosθ

+cosθsin(90°−θ)

cosec(90°−θ)

sinθ

−2sin(90°−θ)cos(90°−θ)=0