Question

If x=k sinAcosB, y = k sinAsinB and z = k cosA, then prove that x2+y2+z2 = k2

Answer 1

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Answer 2

Answer:

Explanation:

ANSWER:-

We are given the following:-

• x=k sinAcosB
• y = k sinAsinB
• z = k cosA

We will directly place the following as

$x^{2} +y^{2} +z^{2}$

$(k sinAcosB)^{2} +( k sinAsinB)^{2} +( k cosA)^{2}$

$= k^{2} sin^{2}Acos^{2}B +k^{2}sin^{2}Asin^{2}B+k^{2}cos^{2}A$

Taking k square common,

$k^{2}(sin^{2}Acos^{2}B+sin^{2}Asin^{2}B+cos^{2}A)$

Now, inside bracket

Taking Sin square A common,

$k^{2}(sin^{2}A(cos^{2}B+sin^{2}B)+cos^{2}A)$

Now we know the Identity

$sin^{2}\theta+cos^{2}\theta=1$

So we can write

$k^{2}(sin^{2}A(1)+cos^{2}A)$

Again, using same identity, we get the final Answer as

$\huge{\boxed{k^{2}}}$

So we have proved that :-

$\boxed{x^{2}+y^{2}+z^{2}=k^{2}}$

Main Identity used here is :-

$sin^{2}\theta+cos^{2}\theta=1$

Other similar Identities are:-

$tan^{2}\theta+1=sec^{2}\theta$

$1+cot^{2}\theta = cosec^{2}\theta$

Complementary Angles:-

• These angles make a sum equal to 90 degree.
• For example in Trigonometry we have 3 angles ( excluding the inverse)

$sin^{2}\theta + cos^{2}\theta=1$

$tan^{2}\theta + 1^=sec^{2}\theta$

$cot^{2}\theta + 1=cosec^{2}\theta$

• These identities are complementary of each other
• Many other identities can be made with these identities.