Question

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Answer 1

cos²A - sin²A = tan²B

_________ [ GIVEN ]

• We have to prove that cos²B - sin²B = tan²A.

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→ cos²A - sin²A = tan²B _______ (eq 1)

We know that

→ sin²A + cos²A = 1

→ cos²A = 1 - sin²A

Put value of cos²A in (eq 1)

→ (1 - sin²A) - sin²A = tan²B

→ 1 - 2sin²A = tan²B

→ - 2sin²A = tan²B - 1

→ 2sin²A = 1 - tan²B

→ sin²A = 1/2(1 - tan²B) _____ (eq 2)

Similarly,

→ sin²A = 1 - cos²A

Put value of sin²A in (eq 1)

→ cos²A - (1 - cos²A) = tan²B

→ cos²A - 1 + cos²A = tan²B

→ 2cos²A - 1 = tan²B

→ 2cos²A = tan²B + 1

→ cos²A = 1/2(tan²B + 1)

Now.. 1 + tan²B = sec²B

So,

→ cos²A = 1/2(sec²B) ______ (eq 3)

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Divide (eq 2) and (eq 3)

→ $\dfrac{ {sin}^{2} A}{ {cos}^{2}A } \: = \: \dfrac{ \dfrac{1}{2} (1 \: - \: {tan}^{2} B)}{ \dfrac{1}{2} {(sec}^{2}B )}$

→ tan²A = $\dfrac{ (1 \: - \: {tan}^{2} B)}{ {(sec}^{2}B )}$

→ tan²A = $\dfrac{1}{ {sin}^{2} B} \: - \: \dfrac{ {tan}^{2} B}{sec ^{2}B }$

Now..

$\dfrac{1}{sinB}$ = cosB

tanB = $\dfrac{sinB}{cosB}$

→ tan²A = cos²B - $\dfrac{ \frac{ {sin}^{2} B}{ {cos}^{2} B} }{ \frac{1}{ {cos}^{2}B } }$

→ tan²A = cos²B - sin²B

___ [ HENCE PROVED ]

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