If $$x=$$$$sin$$−$$1a6+1+$$$$cos$$−$$1a4+1$$−$$tan$$−$$1a2+1$$, a∈R then the value of $$sec2x=$$

Since $$sin$$$$-1is$$ defined on $$-1$$,$$1$$, we must have $$-1$$≤$$a6+1$$≤$$1$$ ⇒$$-2$$≤$$a6$$≤$$0$$ ⇒$$a6=0$$ ⇒$$a=0$$ ∴$$x=sin11+$$$$cos$$$$-11-$$$$tan$$$$-11$$ $$=$$π$$2+0-$$π$$4$$ $$=$$π$$4$$ ∴$$sec2x=sec2$$$$π$$$$4=2$$

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If $x = \sin^{- 1} (a^{6} + 1) + \cos^{- 1} (a^{4} + 1) - \tan^{- 1} (a^{2} + 1), a \in R$ then the value of $\sec^{2} x =$

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Solution

$S i n c e \sin^{- 1} i s d e f i n e d o n [- 1, 1], w e m u s t h a v e - 1 \leq a^{6} + 1 \leq 1 \Rightarrow - 2 \leq a^{6} \leq 0 \Rightarrow a^{6} = 0 \Rightarrow a = 0 ∴ x = \sin^{1} (1) + \cos^{- 1} (1) - \tan^{- 1} (1) = \frac{π}{2} + 0 - \frac{π}{4} = \frac{π}{4} ∴ s e c^{2} x = s e c^{2} (\frac{π}{4}) = 2$

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Introduction to ITF

Remember concepts with our Masterclasses.

If x=sin−1a6+1+cos−1a4+1−tan−1a2+1, a∈R then the value of sec2x=

Since sin-1is defined on -1,1, we must have -1≤a6+1≤1 ⇒-2≤a6≤0 ⇒a6=0 ⇒a=0 ∴x=sin11+cos-11-tan-11 =π2+0-π4 =π4 ∴sec2x=sec2π4=2

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If $x = \sin^{- 1} (a^{6} + 1) + \cos^{- 1} (a^{4} + 1) - \tan^{- 1} (a^{2} + 1), a \in R$ then the value of $\sec^{2} x =$