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Question 56 of 66

Q.If y = tan⁻¹(secx+tanx), then find the value of d²y/dx² at x = π/4.

West Bengal WbchseWest Bengal HS (WBCHSE) Board 2025Subjective· 2mImportance★★★★★
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The expression sec⁡x+tan⁡x\sec x+\tan x simplifies via a standard identity to tan⁡(π/4+x/2)\tan(\pi/4+x/2), so yy is linear in xx.

Using sec⁡x+tan⁡x=1+sin⁡xcos⁡x\sec x+\tan x = \dfrac{1+\sin x}{\cos x}, a standard trigonometric identity gives 1+sin⁡xcos⁡x=tan⁡ ⁣(π4+x2)\dfrac{1+\sin x}{\cos x} = \tan\!\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right).

So:

y=tan⁡−1(sec⁡x+tan⁡x)=π4+x2y = \tan^{-1}(\sec x+\tan x) = \frac{\pi}{4}+\frac{x}{2}

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