1. Find the H.C.F. of 4x2y3 and 6xy2z.
Solution:
The H.C.F. of numerical coefficients = The H.C.F. of 4 and 6.
Since, 4 = 2 × 2 = 22 and 6 = 2 × 3 = 21 × 31
Therefore, the H.C.F. of 4 and 6 is 2
√(4) is equal to 4 raised to the (1/2). 1/2 as an exponent is between 0 and 1 and it means to take the square root.
So, 4^(1/2) = 2 (or -2)
Answer:
see explanation
Step-by-step explanation:
Using the identity
tan²x = sec²x - 1
Consider left side
sec²x + tan²x
= sec²x + sec²x - 1
= 2sec²x - 1
= right side , thus verified
