Prove - tan- - tan 90 - - - - - 90 - - -

tanθ+tan(90^∘ θ)=tanθ+θ =sinθcosθ+cosθsinθ =1sinθcosθ =θθ =(90^∘ θ)θ ...

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

Byju's Answer

Standard X

Mathematics

Trigonometric Identities

Prove : tanθ...

Question

Prove :
$tan θ + tan (90^{\circ} - θ) = sec θ sec (90^{\circ} - θ)$

Open in App

Solution

$tan θ + tan (90^{\circ} - θ) = tan θ + cot θ$
$= \frac{sin θ}{cos θ} + \frac{cos θ}{sin θ}$
$= \frac{1}{sin θ cos θ}$
$= csc θ sec θ$
$= sec (90^{\circ} - θ) sec θ$

Suggest Corrections

Similar questions

Q. Prove the following :

\frac{cos (90^{\circ} - θ) sec (90^{\circ} - θ) tan θ}{cosec (90^{\circ} - θ) sin (90^{\circ} - θ) cot (90^{\circ} - θ)} + \frac{tan (90^{\circ} - θ)}{cot θ} = 2

Prove the following:

(i) $s i n θ s i n (90^{o} - θ) - c o s θ c o s (90^{o} - θ) = 0$
(ii) $\frac{c o s (90^{o} - θ) s e c (90^{o} - θ) t a n θ}{c o s e c (90^{o}) s i n (90^{o} - θ) c o t (90^{o} - θ)} + \frac{t a n (90^{o} - θ)}{c o t θ} = 2$
(iii) $\frac{t a n (90^{o} - A) c o t A}{c o s e c^{2} A} - c o s^{2} A = 0$
(iv) $\frac{c o s (90^{o} - A) s i n (90^{o} - A)}{t a n (90^{o} - A)} = s i n^{2} A$
(v) $s i n (50^{o} + θ) - c o s (40^{o} - θ) + t a n 1^{o} t a n 10^{o} t a n 20^{o} t a n 70^{o} t a n 80^{o} t a n 89^{o} = 1$