Question

1) sinΘ + cosΘ = √2, Find Θ
2) cosecA - cotA = 1/2, Find cosΘ
3) cotΘ + tanΘ = 2 secΘ, FindΘ​

Answer 1

SOLUTION

TO DETERMINE

1) Find Θ when sinΘ + cosΘ = √2

2) Find cosΘ when cosecΘ - cotΘ = 1/2

3) Find Θ when cotΘ + tanΘ = 2 secΘ

EVALUATION

1.

$\sf{ \sin \theta + \cos \theta = \sqrt{2} \: }$

Squaring both sides we get

$\sf{ {\sin}^{2} \theta + {\cos}^{2} \theta +2 \sin \theta \: \cos\theta \: = 2}$

$\implies \sf{ 1 +2 \sin \theta \: \cos\theta \: = 2}$

$\implies \sf{2 \sin \theta \: \cos\theta \: = 1}$

Now

$\sf{ {(\sin \theta - \cos \theta)}^{2} \: }$

$= \sf{ {\sin}^{2} \theta + {\cos}^{2} \theta - 2 \sin \theta \: \cos\theta \: }$

$= \sf{ 1 - 1\: }$

$= \sf{0}$

So

$\sf{ {(\sin \theta - \cos \theta)}^{2} \: } = 0$

$\implies\sf{ {\sin \theta - \cos \theta} \: } = 0$

$\implies\sf{ {\sin \theta = \cos \theta} \: }$

$\implies\sf{ \tan \theta = 1}$

$\displaystyle \boxed{ \sf{ \therefore \: \: \theta = \frac{\pi}{4} } \: \: \: }$

2.

Clearly Cosθ # 0

If Cosθ = 0 then tanθ is undefined

Here it is given that

( Cosecθ - Cotθ ) = (1/2) ......... (1)

We are aware of the identity that

Cosec²θ - Cot²θ = 1

➙( Cosecθ + Cotθ ) ( Cosecθ - Cotθ) = 1

➙( Cosecθ + Cotθ ) ( 1/2) = 1

➙( Cosecθ + Cotθ ) = 2 ......... (2)

Solving Equation 1 and Equation 2 we get

Cosecθ = 5/4 and Cotθ = 3/4

So Cosθ

= Cotθ / Cosecθ

= 3/5

$\boxed{\displaystyle \sf{\cos \theta = \frac{3}{5} }}$

3.

Here

$\displaystyle \sf{ \cot \theta \: + \tan \theta = 2 \sec \theta \: }$

$\implies \displaystyle \sf{ \frac{ \cos \theta }{ \sin \theta } + \frac{ \sin \theta }{ \cos \theta } = \frac{2 }{ \cos\theta } \: }$

$\implies \displaystyle \sf{ \frac{ {\cos}^{2} \theta + {\cos}^{2} \theta }{ \sin \theta \cos \theta} = \frac{2 }{ \cos\theta } \: }$

$\implies \displaystyle \sf{ \frac{ 1}{ \sin \theta \cos \theta} = \frac{2 }{ \cos\theta } \: }$

$\implies \displaystyle \sf{ \frac{ 1}{ \sin \theta } = 2 } \: ( \because \: \: cos \theta \ne \: 0 \: )$

$\implies \displaystyle \sf{ { \sin \theta } = \frac{1}{2} } \:$

$\boxed{\displaystyle \sf{ \therefore \: \: { \theta } = \frac{\pi}{6} } \: }$

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