Question

write the given expression in terms of x and y only tan(sin−1 x + cos−1 y)

Answer 1

let sin\(\theta\)=x and cos\(\alpha\)=y
now the expression is tan(\(\theta+\alpha\))

Answer 2

and then? what is the value of sin? cos?

Answer 3

sin\(\theta\)=x
so cos\(\theta\)=\(\sqrt{1-x^2}\)
now tan(\(\theta+\alpha\)=tan\(\theta\)+tan\(\alpha\)/(1-tan\(\theta\)tan\(\alpha\))

Answer 4

tangent will be x/sqrt of 1-x^2?

Answer 5

yes...good

Answer 6

how come that cos= sqrt of 1-x^2?

Answer 7

do u know this?
sin^2\(\theta\)+cos^2\(\theta\)=1

Answer 8

yes

Answer 9

i get it from there...

Answer 10

but sine function has -1 exponent.

Answer 11

sin\(\theta\)=x
sin^2\(\theta\)=x^2
sin^2θ+cos^2θ=1
or x^2+cos^2\(\theta\)=1
or cos\(\theta\)=\(\sqrt{1-x^2}\)

Answer 12

pls help i cant understand. the -1 exponent? I will disregard it?

Answer 13

if sin\(\theta\)=x
then \(\theta\)=sin^-1x

Answer 14

what will be the equivalent of tan(\[\theta\]

Answer 15

pls help show me the answer

Answer 16

lets do it together...

Answer 17

cosα=y
sinα=?

Answer 18

sin\[\sin \alpha =\sqrt{1-y ^{2}}\]

Answer 19

? am I right?

Answer 20

now tan \(\alpha\)=?

Answer 21

\[\frac{ y }{ \sqrt{} }\]

Answer 22

\[\frac{ \sqrt{1-y ^{2}}}{y }\]

Answer 23

now
tan(θ+α)=tanθ+tanα/(1-tanθtanα)
just put what u get fo tan\(\alpha\) and tan\(\theta\)

Answer 24

I will disregard the -1 exponent of sine?

Answer 25

tanθ+tanα/(1-tanθtanα)
put the tan\(\theta\) and tan\(\alpha\) here

Answer 26

then after i put the given? whats next? simplify?

Answer 27

then simplify