मान लीजिए कि $\sin ^{-1} \frac{12}{13}$ = x, $ \cos ^{-1} \frac{4}{5}$ = y, tan-1 $\frac{63}{16}$ = z इस प्रकार $\sin x=\frac{12}{13}, $ $ \cos y=\frac{4}{5}$ tan z = $\frac{63}{16}$ इसलिए $\cos x=\frac{5}{13}$, sin y = $ \frac{3}{5}$, tan x = $ \frac{12}{5}$ और tan y अब $\tan (x+y)=\frac{\tan x+\tan y}{1-\tan x \tan y}$ = $\frac{\frac{12}{5}+\frac{3}{4}}{1-\frac{12}{5} \times \frac{3}{4}}$ = $-\frac{63}{16}$ अतः tan (x + y) = -tan z अर्थात् tan (x + y) = tan (-z) या tan (x + y) = tan ($\pi$ - z) इसलिए x + y = - z or x + y = $\pi$ - z x, y तथा z धनात्मक हैं, इसलिए x + y $\neq$ - z अतः x + y + z = $\pi$ या sin-1 $\frac{12}{13}$ + cos-1 $ \frac{4}{5}$ + tan-1 $\frac{63}{16}$ = $\pi$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.