Concept understanding — Composite and Sum/Difference Identities of Inverse Trigonometric Functions
These are the working-formula properties (Properties VI–X) for combining or composing inverse trig functions, together with the reference-triangle technique for a raw composite like tan(sin−1x).
Composing a trig function with an unrelated inverse trig function (reference triangle). To evaluate f(g−1(x)) where f=g (e.g. cot(sin−1x)), let θ=g−1(x), build a right triangle encoding θ from the definition of g−1 (e.g. sinθ=x gives opposite =x, hypotenuse =1, so adjacent =1−x2 by Pythagoras — taking the adjacent side non-negative since θ∈[−2π,2π] keeps cosine ≥0 there), then read f(θ) straight off the triangle. This proves, e.g., tan(sin−1x)=1−x2x, −1<x<1.
Property VI (addition/subtraction formulas).
sin−1x+sin−1y=sin−1(x1−y2+y1−x2),if x2+y2≤1 or xy<0
with analogous subtraction forms for sine and cosine. Extending the tangent addition formula to three terms gives tan−1x+tan−1y+tan−1z=tan−1[1−xy−yz−zxx+y+z−xyz]; setting the left side equal to π and taking the tangent of both sides (which is 0) proves the classical identity x+y+z=xyz whenever tan−1x+tan−1y+tan−1z=π.
Property VII (double-angle-style formulas, from setting y=x in Property VI).
Property VIII.sin−1(2x1−x2)=2sin−1x for ∣x∣≤21, and sin−1(2x1−x2)=2cos−1x for 21≤x≤1 (the SAME left side splits into two different right sides depending on which half of [−1,1], i.e. which principal-range piece, x falls in). …
The tangent addition formula collapses the sum to tan−1(21); converting that through the double-angle identity 2tan−1t=cos−11+t21−t2 (valid for t≥0) expresses it as 21cos−1(53), matching option (1).
Step 1. Apply the tangent addition formula. With a=41,b=92: a+b=369+368=3617, ab=362=181, so 1−ab=1817.
Step 2. Compute the ratio.1−aba+b=17/1817/36=3617⋅1718=21.
Step 3. Conclude the sum. Since ab=181<1 and both terms are positive acute angles, tan−141+tan−192=tan−121.
Step 4. Convert using the double-angle identity. For t≥0, 2tan−1t=cos−11+t21−t2. With t=21: 1+1/41−1/4=5/43/4=53, so 2tan−121=cos−153, i.e. tan−121=21cos−153. …