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Prove the following: (1 + tanA · tanB)2 + (tanA − tanB)2 = sec2A · sec2B - Mathematics and Statistics

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प्रश्न

Prove the following:

(1 + tanA · tanB)2 + (tanA − tanB)2 = sec2A · sec2B

बेरीज
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उत्तर

L.H.S. = (1 + tan A · tan B)2 + (tan A − tan B)2

= 1 + 2 tan A · tan B + tan2A tan2B + tan2A – 2 tan A · tan B + tan2B

= 1 + tan2A + tan2B + tan2A tan2B

= 1(1 + tan2A) + tan2B(1 + tan2A)

= (1 + tan2A) (1 + tan2B)

= sec2A sec2B

= R.H.S.

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Fundamental Identities
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पाठ 2: Trigonometry - 1 - MISCELLANEOUS EXERCISE - 2 [पृष्ठ ३४]

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बालभारती Mathematics and Statistics 1 (Arts and Science) [English] Standard 11 Maharashtra State Board
पाठ 2 Trigonometry - 1
MISCELLANEOUS EXERCISE - 2 | Q 10) xiv) | पृष्ठ ३४

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