Question

$\huge\underline{\underline{\boxed{\bf QueStion}}}$

if $\sin( \theta) - \cos( \theta) = \dfrac{1}{2}$ then find the value of $\dfrac{1}{ \sin( \theta) + \cos( \theta) }$

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Answer 1

Solution

Given :-

• $\sin( \theta) - \cos( \theta) = \dfrac{1}{2}$

Find :-

• Value of $\dfrac{1}{ \sin( \theta) + \cos( \theta) }$.

Explanation

We Have,

➠ sin θ - cos θ = 1/2

Squaring both sides

➠ (sin θ - cos θ )² = (1/2)²

➠ sin² θ + cos² θ - 2sin θ.cos θ = 1/4

We Know,

• sin² θ + cos² θ = 1

➠ 1 - 2sin θ.cos θ = 1/4

➠ 2sin θ.cos θ = 1 -1/4

➠ 2sin θ.cos θ = (4 - 1)/3

➠ 2sin θ.cos θ = 3/4

➠ 2sin θ.cos θ = 3/4 ____________(1)

We Know,

• (sin θ + cos θ)² = sin² θ + cos² θ + 2sin θ.cos θ

keep all above Values

➠ (sin θ + cos θ)² = 1 + 3/4

➠ (sin θ + cos θ)² = (4 + 3)/4

➠ (sin θ + cos θ)² = 7/4

Take squarroot of both side

➠ √(sin θ + cos θ)² = √(7/4)

➠ sin θ + cos θ = √7/2

Now, take reciprocal of both side of all terms

➠ 1/(sin θ + cos θ) = 2/√7

Hence

• Value of 1/(sin θ + cos θ) will be = 2/√7

___________________

Answer 2

Given :-

$\sin( \theta) - \cos( \theta) = \dfrac{1}{2}sin(θ−cos(θ)= 21$

Find :-

$Value of \dfrac{1}{ \sin( \theta) + \cos( \theta) } </p><h3>sin(θ)+cos(θ)$

Explanation

We Have,

➠ sin θ - cos θ = 1/2

Squaring both sides

➠ (sin θ - cos θ )² = (1/2)²

➠ sin² θ + cos² θ - 2sin θ.cos θ = 1/4

We Know,

sin² θ + cos² θ = 1

➠ 1 - 2sin θ.cos θ = 1/4

➠ 2sin θ.cos θ = 1 -1/4

➠ 2sin θ.cos θ = (4 - 1)/3

➠ 2sin θ.cos θ = 3/4

➠ 2sin θ.cos θ = 3/4 ____________(1)

We Know,

(sin θ + cos θ)² = sin² θ + cos² θ + 2sin θ.cos θ

keep all above Values

➠ (sin θ + cos θ)² = 1 + 3/4

➠ (sin θ + cos θ)² = (4 + 3)/4

➠ (sin θ + cos θ)² = 7/4

Take squarroot of both side

➠ √(sin θ + cos θ)² = √(7/4)

➠ sin θ + cos θ = √7/2

Now, take reciprocal of both side of all terms

➠ 1/(sin θ + cos θ) = 2/√7

Hence

Value of 1/(sin θ + cos θ) will be = 2/√7

_________________