How many complex numbers have modulus of 5?
2 answers:
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Answer:
Step-by-step explanation:
1.
cot x sec⁴ x = cot x+2 tan x +tan³x
L.H.S = cot x sec⁴x
=cot x (sec²x)²
=cot x (1+tan²x)² [ ∵ sec²x=1+tan²x]
= cot x(1+ 2 tan²x +tan⁴x)
=cot x+ 2 cot x tan²x+cot x tan⁴x
=cot x +2 tan x + tan³x [ ∵cot x tan x =1]
=R.H.S
2.
(sin x)(tan x cos x - cot x cos x)=1-2 cos²x
L.H.S =(sin x)(tan x cos x - cot x cos x)
= sin x tan x cos x - sin x cot x cos x
= sin²x -cos²x
=1-cos²x-cos²x
=1-2 cos²x
=R.H.S
3.
1+ sec²x sin²x =sec²x
L.H.S =1+ sec²x sin²x
= [
]
=1+tan²x
=sec²x
=R.H.S
4.
L.H.S=
= 2 csc x
= R.H.S
5.
-tan²x + sec²x=1
L.H.S=-tan²x + sec²x
= sec²x-tan²x
=
=1