Question

value of sec Square 26 degrees - cot square 64 degrees is

Please answer in a little bit detail ​

Answer 1

Answer:

The result is 1

Step-by-step explanation:

we have

$sec^{2}(26\°)-cot^2(64\°)$

Remember that

$cot^2(64\°)=\frac{cos^2(64\°)}{sin^2(64\°)}$

$sec^{2}(26\°)=\frac{1}{cos^2(26\°)}$

If two angles are complementary ----> A+B=90°

then

cos (A)=sin(B)

In this problem 26° and 64° are complementary angles

therefore

$\frac{1}{cos^2(26\°)}=\frac{1}{sin^2(64\°)}$

substitute

$\frac{1}{sin^2(64\°)}-\frac{cos^2(64\°)}{sin^2(64\°)}$

$\frac{1-cos^2(64\°)}{sin^2(64\°)}$

we have that

$1-cos^2(64\°)=sin^2(64\°)$

substitute

$\frac{sin^2(64\°)}{sin^2(64\°)}=1$