Question

2}(csc(α)−csc(β))2+(sin(α)−sin(β))2=4sin22α−β​
​

Answer 1

Answer:

c s c ( x )=

s i n ( x )

1

\ sin (x) = \ dfrac {1} {\ csc (x)}s i n ( x )=

c s c ( x )

1

\ sec (x) = \ dfrac {1} {\ cos (x)}s e c ( x )=

c o s ( x )

1

\ cos (x) = \ dfrac {1} {\ sec (x)}c o s ( x )=

s e c ( x )

1

\ cot (x) = \ dfrac {1} {\ tan (x)} = \ dfrac {\ cos (x)} {\ sin (x)}c o t ( x )=

t a n ( x )

1

=

s i n ( x )

c o s ( x )

\ tan (x) = \ dfrac {1} {\ cot (x)} = \ dfrac {\ sin (x)} {\ cos (x)}t a n ( x )=

c o t ( x )

1

=

c o s ( x )

s i n ( x )

Answer 2

Answer:

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sorry for