Inverse Trigonometric Functionsप्रतिलोम त्रिकोणमितीय फलनInverse Trigonometric Functions

• sin⁻¹x: domain [−1, 1], principal range [−π/2, π/2].
• cos⁻¹x: domain [−1, 1], principal range [0, π].
• tan⁻¹x: domain R, principal range (−π/2, π/2).
• cot⁻¹x: domain R, principal range (0, π).
• sec⁻¹x: domain R − (−1, 1), principal range [0, π] − {π/2}.
• cosec⁻¹x: domain R − (−1, 1), principal range [−π/2, π/2] − {0}.
• Negative input: sin⁻¹(−x) = −sin⁻¹x, tan⁻¹(−x) = −tan⁻¹x, cosec⁻¹(−x) = −cosec⁻¹x; but cos⁻¹(−x) = π − cos⁻¹x, cot⁻¹(−x) = π − cot⁻¹x, sec⁻¹(−x) = π − sec⁻¹x.
• Method: find the standard angle for the positive value, then apply the negative rule. Example: cos⁻¹(−1/2) = π − π/3 = 2π/3.
• Standard values: sin⁻¹(1/2) = π/6, sin⁻¹(1/√2) = π/4, sin⁻¹(√3/2) = π/3, tan⁻¹1 = π/4, tan⁻¹√3 = π/3, tan⁻¹(1/√3) = π/6.

• sin⁻¹x: प्रांत [−1, 1], मुख्य परिसर [−π/2, π/2]।
• cos⁻¹x: प्रांत [−1, 1], मुख्य परिसर [0, π]।
• tan⁻¹x: प्रांत R, मुख्य परिसर (−π/2, π/2)।
• cot⁻¹x: प्रांत R, मुख्य परिसर (0, π)।
• sec⁻¹x: प्रांत R − (−1, 1), मुख्य परिसर [0, π] − {π/2}।
• cosec⁻¹x: प्रांत R − (−1, 1), मुख्य परिसर [−π/2, π/2] − {0}।
• ऋणात्मक मान: sin⁻¹(−x) = −sin⁻¹x, tan⁻¹(−x) = −tan⁻¹x, cosec⁻¹(−x) = −cosec⁻¹x; परन्तु cos⁻¹(−x) = π − cos⁻¹x, cot⁻¹(−x) = π − cot⁻¹x, sec⁻¹(−x) = π − sec⁻¹x।
• विधि: धनात्मक मान का मानक कोण ज्ञात कीजिए, फिर ऋणात्मक नियम लगाइए। उदाहरण: cos⁻¹(−1/2) = π − π/3 = 2π/3।
• मानक मान: sin⁻¹(1/2) = π/6, sin⁻¹(1/√2) = π/4, sin⁻¹(√3/2) = π/3, tan⁻¹1 = π/4, tan⁻¹√3 = π/3, tan⁻¹(1/√3) = π/6।

• sin⁻¹x: domain [−1, 1], principal range [−π/2, π/2].
• cos⁻¹x: domain [−1, 1], principal range [0, π].
• tan⁻¹x: domain R, principal range (−π/2, π/2).
• cot⁻¹x: domain R, principal range (0, π).
• sec⁻¹x: domain R − (−1, 1), principal range [0, π] − {π/2}.
• cosec⁻¹x: domain R − (−1, 1), principal range [−π/2, π/2] − {0}.
• Negative value: sin⁻¹(−x) = −sin⁻¹x, tan⁻¹(−x) = −tan⁻¹x, cosec⁻¹(−x) = −cosec⁻¹x; lekin cos⁻¹(−x) = π − cos⁻¹x, cot⁻¹(−x) = π − cot⁻¹x, sec⁻¹(−x) = π − sec⁻¹x.
• Method: positive value ka standard angle nikaliye, phir negative wala rule lagaiye. Example: cos⁻¹(−1/2) = π − π/3 = 2π/3.
• Standard values: sin⁻¹(1/2) = π/6, sin⁻¹(1/√2) = π/4, sin⁻¹(√3/2) = π/3, tan⁻¹1 = π/4, tan⁻¹√3 = π/3, tan⁻¹(1/√3) = π/6.

💡 sec⁻¹(−2) is 2π/3, not −π/3: sec⁻¹ never gives a negative angle.sec⁻¹(−2) का मान 2π/3 है, −π/3 नहीं: sec⁻¹ कभी ऋणात्मक कोण नहीं देता।sec⁻¹(−2) ki value 2π/3 hai, −π/3 nahi: sec⁻¹ kabhi negative angle nahi deta.

• √(a² − x²) → put x = a sin θ (or a cos θ). √(a² + x²) → put x = a tan θ. √(x² − a²) → put x = a sec θ.
• √(1 + x), √(1 − x) or (1 − x²)/(1 + x²) → put x = cos 2θ or x = tan θ.
• (cos x − sin x)/(cos x + sin x): divide by cos x to get (1 − tan x)/(1 + tan x) = tan(π/4 − x).
• (3a²x − x³)/(a³ − 3ax²): put x = a tan θ to get tan 3θ.
• After substitution you get tan⁻¹(tan φ) = φ only when φ is in the principal range; then change θ back to x.
• Results to remember: tan⁻¹[x/√(a² − x²)] = sin⁻¹(x/a); cot⁻¹[1/√(x² − 1)] = sec⁻¹x (x > 1); sin(tan⁻¹x) = x/√(1 + x²); cos(sec⁻¹x + cosec⁻¹x) = cos(π/2) = 0.

• √(a² − x²) → x = a sin θ (या a cos θ) रखिए। √(a² + x²) → x = a tan θ। √(x² − a²) → x = a sec θ।
• √(1 + x), √(1 − x) या (1 − x²)/(1 + x²) → x = cos 2θ या x = tan θ रखिए।
• (cos x − sin x)/(cos x + sin x): cos x से भाग देकर (1 − tan x)/(1 + tan x) = tan(π/4 − x)।
• (3a²x − x³)/(a³ − 3ax²): x = a tan θ रखने पर tan 3θ।
• प्रतिस्थापन के बाद tan⁻¹(tan φ) = φ तभी जब φ मुख्य परिसर में हो; फिर θ को x में बदलिए।
• याद रखें: tan⁻¹[x/√(a² − x²)] = sin⁻¹(x/a); cot⁻¹[1/√(x² − 1)] = sec⁻¹x (x > 1); sin(tan⁻¹x) = x/√(1 + x²); cos(sec⁻¹x + cosec⁻¹x) = cos(π/2) = 0।

• √(a² − x²) → x = a sin θ (ya a cos θ) rakhiye. √(a² + x²) → x = a tan θ. √(x² − a²) → x = a sec θ.
• √(1 + x), √(1 − x) ya (1 − x²)/(1 + x²) → x = cos 2θ ya x = tan θ rakhiye.
• (cos x − sin x)/(cos x + sin x): cos x se divide karke (1 − tan x)/(1 + tan x) = tan(π/4 − x).
• (3a²x − x³)/(a³ − 3ax²): x = a tan θ rakhne par tan 3θ.
• Substitution ke baad tan⁻¹(tan φ) = φ tabhi jab φ principal range mein ho; phir θ ko wapas x mein badaliye.
• Yaad rakhiye: tan⁻¹[x/√(a² − x²)] = sin⁻¹(x/a); cot⁻¹[1/√(x² − 1)] = sec⁻¹x (x > 1); sin(tan⁻¹x) = x/√(1 + x²); cos(sec⁻¹x + cosec⁻¹x) = cos(π/2) = 0.

• Main tool: tan⁻¹x + tan⁻¹y = tan⁻¹[(x + y)/(1 − xy)] when xy < 1; tan⁻¹x − tan⁻¹y = tan⁻¹[(x − y)/(1 + xy)] when xy > −1.
• If x > 0, y > 0 and xy > 1, then tan⁻¹x + tan⁻¹y = π + tan⁻¹[(x + y)/(1 − xy)].
• To mix sin⁻¹, cos⁻¹ and tan⁻¹: draw a right triangle and convert everything to tan⁻¹. Example: sin⁻¹(12/13) = tan⁻¹(12/5), cos⁻¹(4/5) = tan⁻¹(3/4).
• Sum of four tan⁻¹ terms: pair them, combine each pair with the formula, then combine the two results.
• Double angle forms: 2tan⁻¹x = tan⁻¹[2x/(1 − x²)] (|x| < 1) = sin⁻¹[2x/(1 + x²)] (|x| ≤ 1) = cos⁻¹[(1 − x²)/(1 + x²)] (x ≥ 0).
• Triple angle: 3sin⁻¹x = sin⁻¹(3x − 4x³), x ∈ [−1/2, 1/2]; 3cos⁻¹x = cos⁻¹(4x³ − 3x), x ∈ [1/2, 1].
• For identities with √(1 + x) and √(1 − x), put x = cos 2θ: √(1 + x) = √2 cos θ, √(1 − x) = √2 sin θ.

• मुख्य उपकरण: tan⁻¹x + tan⁻¹y = tan⁻¹[(x + y)/(1 − xy)] जब xy < 1; tan⁻¹x − tan⁻¹y = tan⁻¹[(x − y)/(1 + xy)] जब xy > −1।
• यदि x > 0, y > 0 तथा xy > 1, तो tan⁻¹x + tan⁻¹y = π + tan⁻¹[(x + y)/(1 − xy)]।
• sin⁻¹, cos⁻¹ तथा tan⁻¹ मिले हों तो समकोण त्रिभुज बनाकर सबको tan⁻¹ में बदलिए। उदाहरण: sin⁻¹(12/13) = tan⁻¹(12/5), cos⁻¹(4/5) = tan⁻¹(3/4)।
• चार tan⁻¹ पदों का योग: जोड़े बनाइए, सूत्र से प्रत्येक जोड़ा मिलाइए, फिर दोनों परिणाम मिलाइए।
• द्विगुण रूप: 2tan⁻¹x = tan⁻¹[2x/(1 − x²)] (|x| < 1) = sin⁻¹[2x/(1 + x²)] (|x| ≤ 1) = cos⁻¹[(1 − x²)/(1 + x²)] (x ≥ 0)।
• त्रिगुण: 3sin⁻¹x = sin⁻¹(3x − 4x³), x ∈ [−1/2, 1/2]; 3cos⁻¹x = cos⁻¹(4x³ − 3x), x ∈ [1/2, 1]।
• √(1 + x) तथा √(1 − x) वाली सर्वसमिकाओं में x = cos 2θ रखिए: √(1 + x) = √2 cos θ, √(1 − x) = √2 sin θ।

• Main tool: tan⁻¹x + tan⁻¹y = tan⁻¹[(x + y)/(1 − xy)] jab xy < 1; tan⁻¹x − tan⁻¹y = tan⁻¹[(x − y)/(1 + xy)] jab xy > −1.
• Agar x > 0, y > 0 aur xy > 1, to tan⁻¹x + tan⁻¹y = π + tan⁻¹[(x + y)/(1 − xy)].
• sin⁻¹, cos⁻¹ aur tan⁻¹ mile hon to right triangle banakar sabko tan⁻¹ mein badaliye. Example: sin⁻¹(12/13) = tan⁻¹(12/5), cos⁻¹(4/5) = tan⁻¹(3/4).
• Chaar tan⁻¹ terms ka jod: pairs banaiye, formula se har pair milaiye, phir dono results milaiye.
• Double angle forms: 2tan⁻¹x = tan⁻¹[2x/(1 − x²)] (|x| < 1) = sin⁻¹[2x/(1 + x²)] (|x| ≤ 1) = cos⁻¹[(1 − x²)/(1 + x²)] (x ≥ 0).
• Triple: 3sin⁻¹x = sin⁻¹(3x − 4x³), x ∈ [−1/2, 1/2]; 3cos⁻¹x = cos⁻¹(4x³ − 3x), x ∈ [1/2, 1].
• √(1 + x) aur √(1 − x) wali identities mein x = cos 2θ rakhiye: √(1 + x) = √2 cos θ, √(1 − x) = √2 sin θ.

💡 Write the condition (xy < 1) next to the formula once; it shows the examiner you know when it works.सूत्र के साथ एक बार शर्त (xy < 1) लिखिए; इससे परीक्षक को दिखता है कि आप जानते हैं सूत्र कब लगता है।Formula ke saath ek baar condition (xy < 1) likhiye; isse examiner ko dikhta hai ki aap jaante hain formula kab lagta hai.