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Find the value of tan22o + tan23o + tan22o.tan23o
- a)-1
- b)1
- c)tan 1°
- d)-tan 1°
- e)5
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
Find the value of tan22o + tan23o + tan22o.tan23oa)-1b)1c)tan 1°d)...
To prove that tan(22o) * tan(23o) = 1, we can use the trigonometric identity:
tan(A) * tan(B) = tan(A+B) - tan(A-B)
Let A = 22o and B = 23o.
Plugging these values into the identity, we get:
tan(22o) * tan(23o) = tan(22o + 23o) - tan(22o - 23o)
Simplifying further:
tan(22o) * tan(23o) = tan(45o) - tan(-1o)
Using the fact that tan(45o) = 1, we can rewrite the expression as:
tan(22o) * tan(23o) = 1 - tan(-1o)
Since tan(-1o) = -tan(1o), we can rewrite the expression as:
tan(22o) * tan(23o) = 1 + tan(1o)
Therefore, the correct answer is:
c) tan 1
tan(A) * tan(B) = tan(A+B) - tan(A-B)
Let A = 22o and B = 23o.
Plugging these values into the identity, we get:
tan(22o) * tan(23o) = tan(22o + 23o) - tan(22o - 23o)
Simplifying further:
tan(22o) * tan(23o) = tan(45o) - tan(-1o)
Using the fact that tan(45o) = 1, we can rewrite the expression as:
tan(22o) * tan(23o) = 1 - tan(-1o)
Since tan(-1o) = -tan(1o), we can rewrite the expression as:
tan(22o) * tan(23o) = 1 + tan(1o)
Therefore, the correct answer is:
c) tan 1
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Community Answer
Find the value of tan22o + tan23o + tan22o.tan23oa)-1b)1c)tan 1°d)...
Tan(A+B)=(tanA+tanB)/(1-tanA*tanB)
Let A=22 degrees,B=23 degrees
Tan(22+23)=(tan22+tan23)/(1-tan22*tan23)
From trigonometry we know tan45=1
Therefore tan(45)=1=(tan22+tan23)/(1-tan22tan23)
=> 1-tan22tan23=tan22+tan23
=> tan22+tan23+tan23tan22 =1
Hence proved
Let A=22 degrees,B=23 degrees
Tan(22+23)=(tan22+tan23)/(1-tan22*tan23)
From trigonometry we know tan45=1
Therefore tan(45)=1=(tan22+tan23)/(1-tan22tan23)
=> 1-tan22tan23=tan22+tan23
=> tan22+tan23+tan23tan22 =1
Hence proved
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Find the value of tan22o + tan23o + tan22o.tan23oa)-1b)1c)tan 1°d)-tan 1°e)5Correct answer is option 'B'. Can you explain this answer?
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