If θ$$1$$,θ$$2$$,θ$$3$$,…,θn are in AP, whose common difference is d, then $$sin$$⁡ dsec⁡θ$$1sec$$⁡θ$$2+$$$$sec$$⁡θ$$2sec$$⁡θ$$3+$$…$$+$$$$sec$$⁡θn−$$1sec$$⁡θn is equal to

Question

If θ$$1$$,θ$$2$$,θ$$3$$,…,θn are in AP, whose common difference is d, then  $$sin$$⁡ dsec⁡θ$$1sec$$⁡θ$$2+$$$$sec$$⁡θ$$2sec$$⁡θ$$3+$$…$$+$$$$sec$$⁡θn−$$1sec$$⁡θn is equal to

Infinity Learn · Accepted Answer

Since,θ$$1$$,θ$$2$$,θ$$3$$,…,θn are in AP⇒ θ$$2$$−θ$$1=$$θ$$3$$−θ$$2=$$…$$=$$θn−θn−$$1=d$$      $$(i)Now$$, taking only first termsin⁡dsec⁡θ$$1sec$$⁡θ$$2=$$$$sin$$⁡dcos⁡θ$$1cos$$⁡θ$$2=$$$$sin$$⁡θ$$2$$−θ$$1cos$$⁡θ$$1cos$$⁡θ$$2=$$$$sin$$⁡θ$$2cos$$⁡θ$$1$$−$$cos$$⁡θ$$2sin$$⁡θ$$1cos$$⁡θ$$1cos$$⁡θ$$2=$$$$sin$$⁡θ$$2cos$$⁡θ$$1cos$$⁡θ$$1cos$$⁡θ$$2$$−$$cos$$⁡θ$$2sin$$⁡θ$$1cos$$⁡θ$$1cos$$⁡θ$$2=$$$$tan$$⁡θ$$2$$−$$tan$$⁡θ$$1Similarly$$, we can solve other terms which will betan⁡θ$$3$$−$$tan$$⁡θ$$2$$,$$tan$$⁡θ$$4$$−$$tan$$⁡θ$$3$$,…∴$$sin$$⁡dsec⁡θ$$1sec$$⁡θ$$2+$$$$sec$$⁡θ$$2sec$$⁡θ$$3+$$…$$+$$$$sec$$⁡θn−$$1sec$$⁡θ$$n=$$$$tan$$⁡θ$$2$$−$$tan$$⁡θ$$1+$$$$tan$$⁡θ$$3$$−$$tan$$⁡θ$$2$$ $$+$$…$$+$$$$tan$$⁡θn−$$tan$$⁡θn−$$1$$ $$=$$−$$tan$$⁡θ$$1+$$$$tan$$⁡θ$$n=$$$$tan$$⁡θn−$$tan$$⁡θ$$1$$

If $θ_{1}, θ_{2}, θ_{3}, \dots, θ_{n}$ are in AP, whose common difference is d, then $\sin d (\sec θ_{1} \sec θ_{2} + \sec θ_{2} \sec θ_{3}$ $+ \dots + \sec θ_{n - 1} \sec θ_{n})$ is equal to

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If θ1,θ2,θ3,…,θn are in AP, whose common difference is d, then sin⁡ dsec⁡θ1sec⁡θ2+sec⁡θ2sec⁡θ3+…+sec⁡θn−1sec⁡θn is equal to

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If $θ_{1}, θ_{2}, θ_{3}, \dots, θ_{n}$ are in AP, whose common difference is d, then $\sin d (\sec θ_{1} \sec θ_{2} + \sec θ_{2} \sec θ_{3}$ $+ \dots + \sec θ_{n - 1} \sec θ_{n})$ is equal to