If 0- lt- lt-90-sec-2- - cosec-2- - tan - - cot -tan - - cot - tan - - cot - tan - - cot -

The correct option is A tan θ + cot θ √(sec^2θ + cosec^2θ) =√((1 + tan^2θ) + (1 + cot^2θ)) =√(tan^2θ + cot^2θ + 2 tanθ cotθ) (∵ tanθ× cotθ =1 ) =√((tanθ + co ...

Byju's Answer

Standard X

Mathematics

Relation between Trigonometric Ratios

If 0^∘ lt;θ...

Question

If $0^{\circ} < θ < 90^{\circ}$ , $\sqrt{s e c^{2} θ + c o s e c^{2} θ}$ =

tan θ + cot θ

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

tan θ - cot θ

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

- tan θ + cot θ

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

- tan θ - cot θ

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

Open in App

Solution

The correct option is A
tan θ + cot θ

$\sqrt{s e c^{2} θ + c o s e c^{2} θ}$
$= \sqrt{(1 + t a n^{2} θ) + (1 + c o t^{2} θ)}$
$= \sqrt{t a n^{2} θ + c o t^{2} θ + 2 t a n θ c o t θ}$
$(∵ t a n θ \times c o t θ = 1)$
$= \sqrt{(t a n θ + c o t θ)^{2}}$
$= t a n θ + c o t θ$

Suggest Corrections

Similar questions

Q. Cot (90° − θ) = ?
(a) cot θ
(b) −cot θ
(c) tan θ
(d) −tan θ

Q. Prove the following :
(viii)

\sqrt{s e c^{2} θ + \cos e c^{2} θ} = \tan θ + c o t θ

S i m p l i f y (i) \frac{c o t (90 + θ) s e c (- θ) \tan (180 θ - θ)}{s e c (360 - θ) \sin (180 + θ) c o t (90 - θ)}

Q. Prove that
(i)

\frac{\cos (2 π + θ) cosec (2 π + θ) \tan (π / 2 + θ)}{\sec (π / 2 + θ) cosθ \cot (π + θ)} = 1

(ii)

\frac{cosec (90 ° + θ) + \cot (450 ° + θ)}{cosec (90 ° - θ) + \tan (180 ° - θ)} + \frac{\tan (180 ° + θ) + \sec (180 ° - θ)}{\tan (360 ° + θ) - \sec (- θ)} = 2

(iii)

\frac{\sin (180 ° + θ) \cos (90 ° + θ) \tan (270 ° - θ) \cot (360 ° - θ)}{\sin (360 ° - θ) \cos (360 ° + θ) cosec (- θ) \sin (270 ° + θ)} = 1

(iv)

\"{\"1 + cotθ - \sec (\frac{π}{2} + θ)\"}\" \"{\"1 + cotθ + \sec (\frac{π}{2} + θ)\"}\" = 2 cotθ

(v)

\frac{\tan (90 ° - θ) \sec (180 ° - θ) \sin (- θ)}{\sin (180 ° + θ) \cot (360 ° - θ) cosec (90 ° - θ)} = 1

Q. find Value of tan(24◦ +θ) – sec(37◦ +θ) – cot (66◦ -θ) + cosec(53◦ -θ) / tan (29◦ +θ) * cot(61◦ -θ)

If 0∘<θ<90∘, √sec2θ+cosec2θ =

The correct option is A tan θ + cot θ √sec2θ+cosec2θ =√(1+tan2θ)+(1+cot2θ) =√tan2θ+cot2θ+2tanθcotθ (∵tanθ×cotθ=1) =√(tanθ+cotθ)2 =tanθ+cotθ

If $0^{\circ} < θ < 90^{\circ}$ , $\sqrt{s e c^{2} θ + c o s e c^{2} θ}$ =

The correct option is A
tan θ + cot θ

$\sqrt{s e c^{2} θ + c o s e c^{2} θ}$
$= \sqrt{(1 + t a n^{2} θ) + (1 + c o t^{2} θ)}$
$= \sqrt{t a n^{2} θ + c o t^{2} θ + 2 t a n θ c o t θ}$
$(∵ t a n θ \times c o t θ = 1)$
$= \sqrt{(t a n θ + c o t θ)^{2}}$
$= t a n θ + c o t θ$