Question: If 3 cot A = 4, check whether (1−tan²A)/(1+tan²A) = cos²A - sin²A
Consider the following △ABC, where m∠B=90°
fig. 1 (drawn from details provided in question)
Important tip: When solving problems of trigonometry, try drawing a right angled triangle with reference to the details given in the question. This can help you visualize and work faster when calculating trigonometric ratios.
Let's start solving this problem by using the given equation 3cotA=4 and find the values for AB, BC and AC.
Weknow∴From△ABC,∴that,3cotA=4cotA=34cotA=BCAB[cotA=Opposite side to ∠AAdjacent side to ∠A]BCAB=34−−−(1)
Hence, from equation (1) we can say AB=4x and BC=3x
Now, that we have the values of all the sides of △ABC, we'll put the values in LHS and RHS and prove that LHS = RHS. Let's start with LHS.
Solving LHS
Firstly finding the value of tanA.
FromtanA∴tanA△ABC,=Adjacent side to m∠AOpposite side to m∠A=ABBC=4x3x=43−−−(2)
LHS=1+tan2A1−tan2A=1+(43)21−(43)2(Putting the values of tan A from equation (2))=1+1691−169=1616+91616−9=1616+91616−9∴LHS=257−−−(3)
Solving RHS
RHS=cos2A−sin2A=(HypotenuseAdjacent side to ∠A)2−(HypotenuseOpposite side to ∠A)2=(ACAB)2−(ACBC)2(from the figure of △ABC)=(5x4x)2−(5x3x)2=25x216x2−25x29x2=25x27x2∴RHS=257−−−(4)
From equation (3) and (4), we can say LHS=RHS
∴1+tan2A1−tan2A=cos2A−sin2A
Hence, proved.
Suggested Notes:
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Suggested Notes:
tan (A + B) = √3 and tan (A -B) = 1/√3 | Trigonometry Numerical
15 cot A = 8 | Find the value of sin A and sec A | Trigonometry Numerical
PQ = 12cm and PR = 13cm | Find tan P - cot R | Trigonometry Numerical
cot θ = 7/8 | Find all other trigonometric ratios | Trigonometry Numerical
Trigonometry Formulas | sin θ, cos θ, tan θ, cosec θ, sec θ, cot θ | Trigonometry Basics
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