বিপরীত ত্রিকোণমিতিক ফাংশনের যোগ বিয়োগ
A+B+C=π, iftan−12=A and tan−13=B, then ∠C=?A+B+C=\pi,\ if\tan^{-1}2=A\ and\ \tan^{-1}3=B,\ then\ ∠C=?A+B+C=π, iftan−12=A and tan−13=B, then ∠C=?
1
2
π2\frac{\pi}{2}2π
π4\frac{\pi}{4}4π
A+B+C=π⇒tan−12+tan−13=π−C⇒tan−1(2+31−2×3)=π−C⇒tan−1(2+31−6)=π−C⇒tan(π−c)=−1⇒−tanC=−1⇒tanC=1∴C=π4 \begin{array}{l}A+B+C=\pi \\ \Rightarrow \tan ^{-1} 2+\tan ^{-1} 3=\pi-C \\ \Rightarrow \tan ^{-1}\left(\frac{2+3}{1-2 \times 3}\right)=\pi-C \\ \Rightarrow \tan ^{-1}\left(\frac{2+3}{1-6}\right)=\pi-C \\ \Rightarrow \tan (\pi-c)=-1 \\ \Rightarrow-\tan C=-1 \\ \Rightarrow \tan C=1 \\ \therefore C=\frac{\pi}{4} \\\end{array} A+B+C=π⇒tan−12+tan−13=π−C⇒tan−1(1−2×32+3)=π−C⇒tan−1(1−62+3)=π−C⇒tan(π−c)=−1⇒−tanC=−1⇒tanC=1∴C=4π
উদ্দীপক-১: f(x)=cosxf(x)=\cos xf(x)=cosx
উদ্দীপক-2: cot−1(1x)+12sec−1(1+y21−y2)+12cosec−1(1+z22z)=π\cot ^{-1}\left(\frac{1}{x}\right)+\frac{1}{2} \sec ^{-1}\left(\frac{1+y^{2}}{1-y^{2}}\right)+\frac{1}{2} \operatorname{cosec}^{-1}\left(\frac{1+z^{2}}{2 z}\right)=\picot−1(x1)+21sec−1(1−y21+y2)+21cosec−1(2z1+z2)=π.
costan−1sincot−1(x)=? \cos \tan ^{-1} \sin \cot ^{-1}(\mathrm{x})=? costan−1sincot−1(x)=?
tan−1x+tan−1y=?\tan ^{-1} x+\tan ^{-1} y = ? tan−1x+tan−1y=?
tan-12+cot-11/3 এর মান কোনটি?
