$3 (8^{0} + 9^{0} + 5^{0}) =_________$

Open in App

Solution

$(8^{0}, 9^{0}, 5^{0})$ is of the form $a^{0}$ .
We know that, $a^{0} = 1$ .
So, $8^{0} = 1, 9^{0} = 1, 5^{0} = 1$ .
Adding all the values,
$8^{0} + 9^{0} + 5^{0} = 1 + 1 + 1$
$= 3$ .
Now, $3 (8^{0} + 9^{0} + 5^{0}) = 3 (3)$
$= 9$ .
Therefore, $3 (8^{0} + 9^{0} + 5^{0})$ $= 9$ .

Suggest Corrections

Similar questions

From the 8.18, write tan x, cot (90

-

y), sec y, sin (90

-

x), cos ec(90

-

y), cos(90

-

x).

Prove that: $cos θ sin θ - \frac{cos (90^{\circ} - θ) cos (90^{\circ} - θ) cos θ}{sec (90^{\circ} - θ)} - \frac{sin (90^{\circ} - θ) sin (90^{\circ} - θ) sin θ}{cosec (90^{\circ} - θ)} + sec (90^{\circ} - θ) = \frac{1}{sin θ}$

Q. Evaluate:
cot0tan(90-0) - (sec(90-0)cosec0 / sin0cos(90-0)+cos0sin(90-0)

Q. Prove that:

(i)

sinθ \cos (90 ° - θ) + \sin (90 ° - θ) cosθ = 1

(ii)

\frac{sinθ}{\cos (90 ° - θ)} + \frac{cosθ}{\sin (90 ° - θ)} = 2

(iii)

\frac{sinθ \cos (90 ° - θ) cosθ}{\sin (90 ° - θ)} + \frac{cosθ \sin (90 ° - θ) sinθ}{\cos (90 ° - θ)}

(iv)

\frac{\cos (90 ° - θ) \sec (90 ° - θ) tanθ}{cosec (90 ° - θ) \sin (90 ° - θ) \cot (90 ° - θ)} + \frac{\tan (90 ° - θ)}{cotθ} = 2

(v)

\frac{\cos (90 ° - θ)}{1 + \sin (90 ° - θ)} + \frac{1 + \sin (90 ° - θ)}{\cos (90 ° - θ)} = 2 cosecθ

(vi)

\frac{\sec (90 ° - θ) cosecθ - \tan (90 ° - θ) cotθ + \cos^{2} 25 ° + \cos^{2} 65 °}{3 \tan 27 ° \tan 63 °} = \frac{2}{3} \"[\"CBSE 2010\"]\"

(vii)

cotθ \tan (90 ° - θ) - \sec (90 ° - θ) cosecθ + \sqrt{3} \tan 12 ° \tan 60 ° \tan 78 ° = 2 \"[\"CBSE 2010\"]\"

Q. Value of sin(90+theta)= sin(90-theta)= cos(90+theta)= cos(90-theta)=

Join BYJU'S Learning Program

380+90+50=_________

80,90,50 is of the form a0.We know that, a0=1.So, 80=1,90=1,50=1.Adding all the values,80+90+50=1+1+1 =3.Now, 380+90+50=33 =9.Therefore, 380+90+50=9.

$3 (8^{0} + 9^{0} + 5^{0}) =_________$

$(8^{0}, 9^{0}, 5^{0})$ is of the form $a^{0}$ .
We know that, $a^{0} = 1$ .
So, $8^{0} = 1, 9^{0} = 1, 5^{0} = 1$ .
Adding all the values,
$8^{0} + 9^{0} + 5^{0} = 1 + 1 + 1$
$= 3$ .
Now, $3 (8^{0} + 9^{0} + 5^{0}) = 3 (3)$
$= 9$ .
Therefore, $3 (8^{0} + 9^{0} + 5^{0})$ $= 9$ .