Evaluate: ${cos}^{2} 17^{o} - {sin}^{2} 73^{o}$

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!

2

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

3

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

Open in App

Solution

The correct option is A $0$
${cos}^{2} (17^{0}) - {sin}^{2} (73^{0})$

$= {sin}^{2} (90^{0} - 17^{0}) - {sin}^{2} (73^{0})$ .......... $(cos θ = sin (90^{0} - θ))$

$= {sin}^{2} (73^{0}) - {sin}^{2} (73^{0})$
$= 0$

Suggest Corrections

Similar questions

\frac{{sin}^{2} 63^{o} + {sin}^{2} 27^{o}}{{cos}^{2} 17^{o} + {cos}^{2} 73^{o}}

\frac{{sin}^{2} 63^{o} + {sin}^{2} 27^{o}}{{cos}^{2} 17^{o} + {cos}^{2} 73^{o}}

1)

\frac{s i n^{2} 63^{o} + s i n^{2} 27^{o}}{c o s^{2} 17^{o} + c o s^{2} 73^{o}}

2)

s i n 25^{o} c o s 65^{o} + c o s 25^{o} s i n 65^{o}

\frac{{sin}^{2} 63^{0} + {sin}^{2} 27^{o}}{{cos}^{2} 17^{o} + {cos}^{2} 73^{o}}

Evaluate the determinants.

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

Join BYJU'S Learning Program