বিপরীত ত্রিকোণমিতিক ফাংশনের যোগ বিয়োগ
সমাধান কর।
cos−1x−sin−1x=sin−1(1−x)\cos^{-1}x-\sin^{-1}x=\sin^{-1}\left(1-x\right)cos−1x−sin−1x=sin−1(1−x)
6,9
3,4
0,1/2
9,0
sin−1x+sin−1(1−x)=cos−1x⇒sin−1x+sin−1(1−x)=π2−sin−1x⇒sin−1(1−x)=π2−2sin−1x…(i) \begin{array}{l} \sin ^{-1} x+\sin ^{-1}(1-x)=\cos ^{-1} x \\ \Rightarrow \sin ^{-1} x+\sin ^{-1}(1-x)=\frac{\pi}{2}-\sin ^{-1} x \\ \Rightarrow \sin ^{-1}(1-x)=\frac{\pi}{2}-2 \sin ^{-1} x…(i) \end{array} sin−1x+sin−1(1−x)=cos−1x⇒sin−1x+sin−1(1−x)=2π−sin−1x⇒sin−1(1−x)=2π−2sin−1x…(i)
Let sin−1x=y \sin ^{-1} \mathrm{x}=\mathrm{y} sin−1x=y
⇒x=siny \Rightarrow \mathrm{x}=\sin \mathrm{y} ⇒x=siny
Therefore, from (i), we get
sin−1(1−x)=π2−2y⇒1−x=sin(π2−2y)⇒1−x=cos2y⇒1−x=1−2sin2y⇒1−x=1−2x2⇒2x2−x=0⇒x(2x−1)=0⇒x=0,12 \begin{array}{l} \sin ^{-1}(1-x)=\frac{\pi}{2}-2 y \\ \Rightarrow 1-x=\sin \left(\frac{\pi}{2}-2 y\right) \\ \Rightarrow 1-x=\cos 2 y \\ \Rightarrow 1-x=1-2 \sin ^{2} y \\ \Rightarrow 1-x=1-2 x^{2} \\ \Rightarrow 2 x^{2}-x=0 \\ \Rightarrow x(2 x-1)=0 \\ \Rightarrow x=0, \frac{1}{2} \end{array} sin−1(1−x)=2π−2y⇒1−x=sin(2π−2y)⇒1−x=cos2y⇒1−x=1−2sin2y⇒1−x=1−2x2⇒2x2−x=0⇒x(2x−1)=0⇒x=0,21
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.
tan−1x+tan−1y=?\tan ^{-1} x+\tan ^{-1} y = ? tan−1x+tan−1y=?
f(x)=cot(π2−x) এবং g(x)=sin−1x f(x)=\cot \left(\frac{\pi}{2}-x\right) \text { এবং } g(x)=\sin ^{-1} x f(x)=cot(2π−x) এবং g(x)=sin−1x
sin cot−1tan cos−1x\sin\ \cot^{-1}\tan\ \cos^{-1}xsin cot−1tan cos−1x এর মান কত?
