The value of cos-sin-1cos-sin-1-2 sin-cos- is

The correct option is A 0(cosθ+sinθ+1)(cosθ+sinθ 1) 2sinθcosθ nbsp;=(cosθ+sinθ)^2 2sinθcosθ 1 =cos^2θ+sin^2θ 1 =0 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios Using Right Angled Triangle

The value of ...

Question

The value of $(cos θ + sin θ + 1) (cos θ + sin θ - 1) - 2 sin θ cos θ$ is

0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 sin θ

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 cos θ

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $0$
$(cos θ + sin θ + 1) (cos θ + sin θ - 1) - 2 sin θ cos θ = (cos θ + sin θ)^{2} - 2 sin θ cos θ - 1 = {cos}^{2} θ + {sin}^{2} θ - 1 = 0$

Suggest Corrections

Similar questions

Q. Evaluate

{(\begin{matrix} cos θ + sin θ & \sqrt{2} sin θ - \sqrt{2} sin θ & cos θ - sin θ \end{matrix})}^{n}

Prove the following trigonometric identities.
(i) $\frac{1 + cos θ + sin θ}{1 + cos θ - sin θ} = \frac{1 + sin θ}{cos θ}$

(ii) $\frac{sin θ - cos θ + 1}{sin θ + cos θ - 1} = \frac{1}{sec θ - tan θ}$

(iii) $\frac{cos θ - sin θ + 1}{cos θ + sin θ - 1} = cosec θ + cot θ$

(iv) $(sin θ + cos θ) (tan θ + cot θ) = sec θ + cosec θ$

$Solve the following equations : (i) s i n θ + c o s θ = \sqrt{2} (i i) \sqrt{3} c o s θ + s i n θ = 1 (i i i) s i n θ + c o s θ = 1 (i v) c o s e c θ = 1 + c o s θ (v) (\sqrt{3} - 1) c o s θ + (\sqrt{3} + 1) s i n θ = 2$

Q. Assertion :Let

A (θ) = [\begin{matrix} cos θ + sin θ & \sqrt{2} sin θ - \sqrt{2} sin θ & cos θ - sin θ \end{matrix}]

A (π / 3)^{3} = - I

Reason:

A (θ) A (ϕ) = A (θ + ϕ)

Q. The numerical value of

cosec θ [\frac{1 - cos θ}{sin θ} + \frac{sin θ}{1 - cos θ}] - 2 {cot}^{2} θ

Join BYJU'S Learning Program

Explore more

Trigonometric Ratios Using Right Angled Triangle

Standard XII Mathematics

Join BYJU'S Learning Program

The value of (cosθ+sinθ+1)(cosθ+sinθ−1)−2sinθcosθ is

The correct option is A 0(cosθ+sinθ+1)(cosθ+sinθ−1)−2sinθcosθ=(cosθ+sinθ)2−2sinθcosθ−1=cos2θ+sin2θ−1=0

The value of $(cos θ + sin θ + 1) (cos θ + sin θ - 1) - 2 sin θ cos θ$ is

The correct option is A $0$
$(cos θ + sin θ + 1) (cos θ + sin θ - 1) - 2 sin θ cos θ = (cos θ + sin θ)^{2} - 2 sin θ cos θ - 1 = {cos}^{2} θ + {sin}^{2} θ - 1 = 0$