If $$a1$$,$$a2$$,$$a3$$,.......anis an AP with common difference d, then tantan−$$1d1+a1a2+$$$$tan$$−$$1d1+a2a3+$$....$$+$$$$tan$$−$$1d1+an$$−$$1an$$ is equal to

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If  $$a1$$,$$a2$$,$$a3$$,.......anis an AP with common difference d, then tantan−$$1d1+a1a2+$$$$tan$$−$$1d1+a2a3+$$....$$+$$$$tan$$−$$1d1+an$$−$$1an$$ is equal to

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We have $$tan$$−$$1a2$$−$$a11+a1a2+$$$$tan$$−$$1a3$$−$$a21+a2a3+$$......$$+$$$$tan$$−$$1an$$−an−$$11+an$$−$$1an$$  ∵$$d=an-an-1$$ ∀n                           $$=($$$$tan$$−$$1a2$$−$$tan$$−$$1a1)+($$$$tan$$−$$1a3$$−$$tan$$−$$1a2)+$$......$$+($$$$tan$$−$$1an$$−$$tan$$−$$1an$$−$$1)$$                            $$=$$$$tan$$−$$1an$$−$$tan$$−$$1a1=$$$$tan$$−$$1an$$−$$a11+ana1$$                             $$=$$$$tan$$−$$1(n$$−$$1)d1+a1an$$∴tantan−$$1d1+a1a2+$$$$tan$$−$$1d1+a2a3+$$....$$+$$$$tan$$−$$1d1+an$$−$$1an=(n$$−$$1)d1+a1an$$

If $a_{1}, a_{2}, a_{3}, ....... a_{n}$ is an AP with common difference d, then $\tan [\tan^{- 1} (\frac{d}{1 + a_{1} a_{2}}) + \tan^{- 1} (\frac{d}{1 + a_{2} a_{3}}) + .... + \tan^{- 1} (\frac{d}{1 + a_{n - 1} a_{n}})]$ is equal to

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If a1,a2,a3,.......anis an AP with common difference d, then tantan−1d1+a1a2+tan−1d1+a2a3+....+tan−1d1+an−1an is equal to

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If $a_{1}, a_{2}, a_{3}, ....... a_{n}$ is an AP with common difference d, then $\tan [\tan^{- 1} (\frac{d}{1 + a_{1} a_{2}}) + \tan^{- 1} (\frac{d}{1 + a_{2} a_{3}}) + .... + \tan^{- 1} (\frac{d}{1 + a_{n - 1} a_{n}})]$ is equal to