Explain the method used.
Class 10th, Rd sharma Trigonometric identities.
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Solution:
Given,
3cot A = 4
⇒ cot A = 4/3
By definition,
tan A = 1/ Cot A = 1/ (4/3)
⇒ tan A = 3/4
Thus,
Base side adjacent to ∠A = 4
Perpendicular side opposite to ∠A = 3
In ΔABC, Hypotenuse is unknown.
Thus, by applying Pythagoras theorem in ΔABC,
We get
AC2 = AB2 + BC2
AC2 = 42 + 32
AC2 = 16 + 9
AC2 = 25
AC = √25
AC = 5
Hence, hypotenuse = 5
Now, we can find that
sin A = opposite side to ∠A/ Hypotenuse = 3/5
And,
cos A = adjacent side to ∠A/ Hypotenuse = 4/5
Taking the LHS,
Thus, LHS = 7/25
Now, taking RHS,
Solution:
Given,
tan θ = a/b
And we know by definition that
tan θ = opposite side/ adjacent side
Thus, by comparison,
Opposite side = a and adjacent side = b
To find the hypotenuse, we know that by Pythagoras theorem that
Hypotenuse2 = opposite side2 + adjacent side2
⇒ Hypotenuse = √(a2 + b2)
So, by definition
sin θ = opposite side/ Hypotenuse
sin θ = a/ √(a2 + b2)
And,
cos θ = adjacent side/ Hypotenuse
cos θ = b/ √(a2 + b2)
Now,
After substituting for cos θ and sin θ, we have
∴
Hence, proved.