If cos theeta +sin theeta=underroot2 cos theeta,show that cos theeta -sin theeta=underroot 2 sin theeta
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heya☺
it's too easy .
cos¢+sin¢=√2(given)--------1)
(cos¢-sin¢)^2+(cos¢+sin¢)^2 【adding cos¢-sin¢)^2
cos^2¢+sin^2¢-2sin¢.cos¢+cos^2¢+sin^2¢+2cos¢●sin¢
(cos¢+sin¢)^2+(cos¢-sin¢)^2=2
now,from 1)putting cos¢+sin¢=√2
so,we get. ,,cos¢-sin¢=0
hope it help you.
@rajukumar .☺☺
it's too easy .
cos¢+sin¢=√2(given)--------1)
(cos¢-sin¢)^2+(cos¢+sin¢)^2 【adding cos¢-sin¢)^2
cos^2¢+sin^2¢-2sin¢.cos¢+cos^2¢+sin^2¢+2cos¢●sin¢
(cos¢+sin¢)^2+(cos¢-sin¢)^2=2
now,from 1)putting cos¢+sin¢=√2
so,we get. ,,cos¢-sin¢=0
hope it help you.
@rajukumar .☺☺
Answered by
0
Answer:
here is ur required answer
Step-by-step explanation:
Sin θ+cos θ=√2
Now we have to Square on both sides
={sin θ+cos θ}²=√2²
={sin² θ+cos ² θ}+2 sin
θ cos θ=2
=1+sin θ cos θ=2
⇒sin θ cos θ=1/2
Now,
Tan θ+cot θ=sin θ/cos θ+cos θ/sin θ
={sin² θ+cos² θ} /sin θ cos θ
=1/ {1/2} = 2
Hence,the answer is "2".
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