CameraIcon

SearchIcon

MyQuestionIcon

1

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

Allied Angles

If θ=178°, th...

Question

If $θ = 178 °$ , then the value of $sin θ \sqrt{1 + {cot}^{2} θ} + cos θ \sqrt{1 + {tan}^{2} θ}$ is

A

2

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

B

- 2

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

C

0

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

D

1

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

Open in App

Solution

The correct option is C $0$
Given expression $sin θ | cosec θ | + cos θ | sec θ | = sin θ cosec θ + cos θ (- sec θ) (∵ θ \in Q_{2} ∴ cosec θ > 0, sec θ < 0) = 1 - 1 = 0$

Suggest Corrections

thumbs-up

0

Similar questions

\frac{sin θ - cos θ + 1}{sin θ + cos θ - 1} = \frac{1}{sec θ - tan θ}

, using the identity

{sec}^{2} θ = 1 + {tan}^{2} θ

Q. Prove the following.

(1) secθ (1 – sinθ) (secθ + tanθ) = 1

(2) (secθ + tanθ) (1 – sinθ) = cosθ

(3) sec²θ + cosec²θ = sec²θ × cosec²θ

(4) cot²θ – tan²θ = cosec²θ – sec²θ

(5) tan⁴θ + tan²θ = sec⁴θ – sec²θ

(6)

\frac{1}{1 - \sin θ} + \frac{1}{1 + \sin θ} = 2 \sec^{2} θ

(7) sec⁶x – tan⁶x = 1 + 3sec²x × tan²x

(8)

\frac{\tan θ}{s e c θ + 1} = \frac{s e c θ - 1}{\tan θ}

\frac{\tan^{3} θ - 1}{\tan θ - 1} = \sec^{2} θ + \tan θ

\frac{\sin θ - \cos θ + 1}{\sin θ + \cos θ - 1} = \frac{1}{\sin θ - \tan θ}

\frac{sin θ + cos θ}{sin θ - cos θ} + \frac{sin θ - cos θ}{sin θ + cos θ} = \frac{1}{{sin}^{2} θ - {cos}^{2} θ} = \frac{{sec}^{2} θ}{{tan}^{2} θ - 1}

Say true or false.

cos θ = \frac{1}{2}

then the value of

\frac{2 sec θ}{1 + {tan}^{2} θ}

{cot}^{2} θ = \frac{7}{8}

0 < θ < 90^{o}

, then the value of

\frac{(1 + sin θ) (1 - sin θ)}{(1 + cos θ) (1 - cos θ)}

View More

Join BYJU'S Learning Program

explore_more_icon

Explore more

Standard XIII Mathematics

Join BYJU'S Learning Program

CrossIcon