On the right side we have st which is the same as s*t or s times t
We want to undo the multiplication that is happening to s, so we have to divide both sides by t
d = s*t
d/t = s*t/t ..... divide both sides by t
d/t = s
s = d/t
Answer: Choice A)
Answer:
8
Step-by-step explanation:
use pythagoras theorem
a^2+b^2=c^2
a^2+6^2=10^2
a^2=10^2-6^2
a^2=100-36
a^2=64
a=square root of 64
a=8
Answer:
Step-by-step explanation:
(This can be simplified, because when you divide expressions with exponents that have the same bases, you can subtract the exponents from the one with the corresponding base)
= 
(Simplify)
=
Answer:
b)(b²-a²)
Step-by-step explanation:
a cotθ + b cosecθ =p
b cotθ + a cosecθ =q
Now,
p²- q²
=(a cotθ + b cosecθ)² - (b cotθ + a cosecθ)² [a²-b²=(a+b)(a-b)]
=(acotθ+bcosecθ + bcotθ+ acosecθ) (a cotθ + bcosecθ -bcotθ-acosecθ)
={a(cotθ+cosecθ)+b(cotθ+cosecθ)} {a (cotθ-cosecθ)+b (cosecθ-cotθ)}
={a(cotθ+cosecθ)+b(cotθ+cosecθ)} [a (cotθ-cosecθ) + {- b (cotθ-cosecθ)} ]
={a(cotθ+cosecθ)+b(cotθ+cosecθ)} {a (cotθ-cosecθ) - b (cotθ-cosecθ)}
={(cotθ+cosecθ)(a+b)} {(cotθ-cosecθ) (a-b)}
=(cotθ+cosecθ) (a+b) (cotθ-cosecθ) (a-b)
=(cotθ+cosecθ) (cotθ-cosecθ) (a+b) (a-b)
= (cot²θ-cosec²θ) (a²-b²) [(a+b) (a-b)= (a²-b²)]
= -1 . (a²-b²) [ 1+cot²θ=cosec²θ ; ∴cot²θ-cosec²θ=-1]
=(b²-a²)
