Question

Question:-

sin0 + cos0 = √2 (where 0 ° <0 <90 °) and the value is of 0​


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Answer 1

Answer:

LET THETA BE “ X ”

★ GIVEN ★

sin \: x \: + \: cos \: x \: = \sqrt{2}sinx+cosx=2

NOW

BY HIT AND TRIAL METHOD

LET X = 45°

HENCE ,

sin \: 45 \: = \frac{1}{ \sqrt{2} }sin45=21

cos \: 45 = \frac{1}{ \sqrt{2} }cos45=21

NOW

\frac{1}{ \sqrt{2} } + \frac{1}{ \sqrt{2} } = \sqrt{2}21+21=2

\frac{2}{ \sqrt{2} } = \sqrt{2}22=2

HENCE

\frac{ (\sqrt{2} ) ( \sqrt{2}) }{ \sqrt{2} } = \sqrt{2}2(2)(2)=2

NOW

\sqrt{2} = \sqrt{2}2=2

HENCE VALUE OF “ X ” = 45°

OR VALUE OF THETA = 45°

Answer 2

Answer:

0=π/4....

Step-by-step explanation:

because...sin45+cos45=>>>>>1/√2+1/√2=>>>>>>√2