$sin 25^{0} + cos 115^{0}$ = _______

1

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

0

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

- 1

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

\sqrt{3} / 2

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

Open in App

Solution

The correct option is B $0$
We know that $cos (90^{0} + θ) = - sin θ$
So, $sin 25^{0} + cos 115^{0}$
$= sin 25^{0} + cos (90^{0} + 25^{0})$
$= sin 25^{0} - sin 25^{0} = 0$

Suggest Corrections

Similar questions

{sin}^{2} 5^{0} + {sin}^{2} 10^{0} + {sin}^{2} 15^{0} + . . . + {sin}^{2} 90^{0} = a \frac{1}{2} .

a

∣ ∣ ∣ ∣ \begin{matrix} 3 & - 1 & - 2 0 & 0 & - 1 3 & - 5 & - 0 \end{matrix} ∣ ∣ ∣ ∣

∣ ∣ ∣ ∣ \begin{matrix} 3 & - 4 & 5 1 & 1 & - 2 2 & 3 & 1 \end{matrix} ∣ ∣ ∣ ∣

∣ ∣ ∣ ∣ \begin{matrix} 0 & 1 & 2 - 1 & 0 & - 3 - 2 & 3 & 0 \end{matrix} ∣ ∣ ∣ ∣

∣ ∣ ∣ ∣ \begin{matrix} 2 & - 1 & - 2 0 & 2 & - 1 3 & - 5 & 0 \end{matrix} ∣ ∣ ∣ ∣

\sqrt{3} x = 2

\frac{x}{3} = 7 - \frac{y}{5}

(1) 2² + 3²

(2) 3² − 2²

(3) 2² × 3³

(4) (−2)² × (1)²

(5) 1² + 5²

(6) 1² + 2⁰

(7) 7⁰ + 1²

(8) 6¹ − 4⁰

(9) 10⁰ − 2⁰

(10) 5¹× 3¹

I. The centroid of the triangle formed by $(2, 3, - 1), (5, 6, 3), (2, - 3, 1)$ is	a) $(2, 2, 2)$
II. The circumcentre of the triangle formed by $(1, 2, 3), (2, 3, 1), (3, 1, 2)$ is	b) $(3, 1, 4)$
III. The orthocenter of the triangle formed by $(2, 1, 5), (3, 2, 3), (4, 0, 4)$ is	c) $(1, 1, 0)$
IV. The incentre of the triangle formed by $(0, 0, 0), (3, 0, 0), (0, 4, 0)$ is	d) $(3, 2, 1)$
	e) $(0, 0, 0)$

Join BYJU'S Learning Program