Exercise 4.1
| Ex-4.1 |
| Q1-Q5 |
| Q6-Q11 |
| Q12-Q19 |
Question 6
Prove that(i) $tan^{-1} x+cot^{-1} y=tan^{-1} \dfrac{xy+1}{y-x}$
Sol :
L.H.S=tan-1 x + cot-1 y
∵$\left[cot^{-1} x =tan^{-1} \dfrac{1}{x}\right]$
=$tan^{-1} x+tan^{-1}\dfrac{1}{y}$
=$tan^{-1} \left(\dfrac{x+\dfrac{1}{y}}{1-x\times \dfrac{1}{y}}\right)$
=$tan^{-1} \left( \dfrac{\dfrac{xy+1}{y}}{\dfrac{y-x}{y}} \right)$
=$tan^{-1} \dfrac{xy+1}{y-x}$ R.H.S
(ii) tan-1 x + cot-1(1+x) = tan-1(1+x+x2)
Sol :
L.H.S= tan-1 x+cot-1 (1+x)
=$tan^{-1} x+tan^{-1} \dfrac{1}{1+x} $
=$tan^{-1} \left(\dfrac{x+\dfrac{1}{x}}{1-x\times \dfrac{1}{1+x}}\right)$
=$tan^{-1} \left(\dfrac{\dfrac{x(1+x)+1}{1+x}}{\dfrac{1+x-x}{1+x}}\right)$
=$tan^{-1} \dfrac{x+x^2+1}{1}$
= tan-1(1+x+x2) R.H.S
Proved
(iii) $tan^{-1} \dfrac{1}{x+y} + tan^{-1} \dfrac{1}{x^2+xy+1}=cot^-{1} x$
Sol :
L.H.S=$tan^{-1} \dfrac{1}{x+y}+tan^{-1} \dfrac{y}{x+xy+1}$
=$tan^{-1} \left(\dfrac{\dfrac{1}{x+y}+\dfrac{y}{x^2+xy+1}}{1-\dfrac{1}{x+y}\times \dfrac{y}{x^2+xy+1}}\right)$
=$tan^{-1} \dfrac{\left(\dfrac{x^2+xy+1+xy+y^2}{(x+y)(x^2+xy+1)}\right)}{\left(\dfrac{(x+y)(x^2+xy+1)-y}{(x+y)(x^2+xy+1)}\right)}$
=$tan^{-1} \dfrac{x^2+xy+1+xy+y^2}{x^3+xy^2+x+yx^2+xy^2+y-y}$
=$tan^{-1} \dfrac{(x^2+y^2+2xy+1)}{(x^3+xy^2+2x^2y+x)}$
=$tan^{-1} \dfrac{(x^2+y^2+2xy+1)}{(x(x^2+y^2+2xy+1))}$
=$tan^{-1} \dfrac{1}{x}=cot^{-1} x$R.H.S
proved
(iv) 2 cot-1 5+cot-1 7+2 cot-18=π/4
Sol :
L.H.S=2 cot-1 5 + cot-17+2 cot-18
=2 cot-1 5 +2 cot-18+ cot-17
=2(cot-1 5+cot-1 8)+cot-1 7
=$\left(tan^{-1} \dfrac{1}{5}+tan^{-1} \dfrac{1}{8}\right)+tan^{-1} \dfrac{1}{7}$
=$2\left(tan^{-1} \dfrac{\frac{1}{5}+\frac{1}{8}}{1-\frac{1}{5}\times \frac{1}{8}}\right)+tan^{-1} \dfrac{1}{7}$
=$2 tan^{-1}\left(\dfrac{\dfrac{13}{40}}{\dfrac{39}{40}}\right)+tan^{-1} \dfrac{1}{7}$
=$2 tan^{-1} \dfrac{1}{3}+tan^{-1} \dfrac{1}{7}$
=$tan^{-1} \dfrac{2\times \dfrac{1}{3}}{1-\dfrac{1}{9}}+tan^{-1} \dfrac{1}{7}$
=$tan^{-1} \left(\dfrac{\frac{2}{9}}{\frac{8}{9}}\right)+tan^{-1} \dfrac{1}{7}$
=$tan^{-1} \left(\dfrac{3}{4}\right)+tan^{-1} \dfrac{1}{7}$
=$tan^{-1} \left(\dfrac{\dfrac{3}{4}+\dfrac{1}{7}}{1-\dfrac{3}{4}\times\dfrac{1}{7}}\right)$
=$tan^{-1} \dfrac{\dfrac{25}{28}}{\dfrac{25}{28}}=tan^{-1} 1$
=$tan^{-1} tan \dfrac{\pi}{4}= \dfrac{\pi}{4}$ R.H.S
Proved
(v) $tan^{-1} \left(\dfrac{x}{8}\right)-tan^{-1}\left(\dfrac{x-y}{x+y}\right)=\dfrac{\pi}{4}$
Sol :
L.H.S=$tan^{-1} \left(\dfrac{x}{y}\right)-tan^{-1} \left(\dfrac{x-y}{x+y}\right)$
=$tan^{-1} \left\{\dfrac{\frac{x}{y}-\frac{x-y}{x+y}}{1+\frac{x}{y} \times \frac{x-y}{x+y}}\right\}$
=$tan^{-1} \dfrac{\left(\dfrac{x(x+y)-y(x-y)}{y(x+y)}\right)}{\left(\dfrac{y(x+y)+x(x-y)}{y(x+y)}\right)}$
=$tan^{-1} \dfrac{[x^2+xy-yx+y^2]}{[yx+y^2+x^2-yx]}$
=$tan^{-1} \dfrac{[x^2+y^2]}{[x^2+y^2]}=tan^{-1} 1$
=$tan^{-1} tan \dfrac{\pi}{4}=\dfrac{\pi}{4}$ R.H.S
Question 7
Prove that :(i) $tan^{-1} \dfrac{a-b}{1+ab}+tan^{-1} \dfrac{b-c}{1+bc}+tan^{-1} \dfrac{c-a}{1+ca}=0$ ab>-1 , bc>-1 , ca>-1
Sol :
L.H.S=$tan^{-1} \dfrac{a-b}{1+ab}+tan^{-1} \dfrac{b-c}{1+bc}+tan^{-1} \dfrac{c-a}{1+ca}$
=tan-1 a-tan-1 b+tan-1 b-tan-1 c+tan-1 c-tan-1 a
=tan-1 a-tan-1 a-tan-1 b+tan-1 b-tan-1 c+tan-1 c
=0 =R.H.S
proved
(ii) $tan^{-1} \dfrac{a^3-b^3}{1+a^3b^3}+tan^{-1} \dfrac{c^3-a^3}{1+b^3c^3}+tan^{-1} \dfrac{c^3-a^3}{1+c^3a^3}=0$
Sol :
L.H.S=$tan^{-1} \dfrac{a^3-b^3}{1+a^3b^3}+tan^{-1} \dfrac{c^3-a^3}{1+b^3c^3}+tan^{-1} \dfrac{c^3-a^3}{1+c^3a^3}$
=tan-1 a3-tan-1 b3+tan-1 b3-tan-1 c3+tan-1 c3-tan-1 a3
=tan-1 a3-tan-1 a3-tan-1 b3+tan-1 b3-tan-1 c3+tan-1 c3
=0 R.H.S
proved
Question 8
(i) $sin^{-1} \dfrac{3}{5}+sin^{-1} \dfrac{8}{17}=sin^{-1} \dfrac{77}{85}$Sol :
L.H.S=$=\sin ^{-1} \frac{3}{5}+\sin ^{-1} \frac{8}{17}$
$=\sin ^{-1}\left[\frac{3}{5} \cdot \sqrt{1-\frac{8^{2}}{17^{2}}}+\frac{8}{17} \cdot \sqrt{1-\left(\frac{3}{5}\right)^{2}}\right]$
=$\sin ^{-1}\left[\frac{3}{5} \times \frac{\sqrt{17^{2}-8^{2}}}{17}+\frac{8}{17} \cdot \frac{\sqrt{5^{2}-3^{2}}}{5}\right]$
=$\sin ^{-1}\left[\frac{3}{5} \times \frac{\sqrt{289-64}}{17}+\frac{8}{17} \cdot \frac{\sqrt{25-9}}{5}\right]$
$=\sin ^{-1}\left[\frac{3}{5} \times \frac{\sqrt{225}}{17}+\frac{8}{17} \cdot \frac{\sqrt{16}}{5}\right]$
$=\sin ^{-1}\left[\frac{3}{5} \times \frac{15}{17}+\frac{8}{17} \times \frac{4}{5}\right]$
$=\sin ^{-1}\left[\frac{45}{85}+\frac{32}{85}\right]$
$=\sin ^{-1}\left[\frac{77}{85}\right]$=R.H.S proved
(ii) $\cos ^{-1} \frac{3}{5}+\cos ^{-1} \frac{12}{13}+\cos ^{-1} \frac{63}{65}=\frac{\pi}{2}$
Sol :
L.H.S$=\cos ^{-1} \frac{3}{5}+\cos ^{-1} \frac{12}{13}+\cos ^{-1} \frac{63}{65}$
$=\cos ^{-1}\left(\frac{3}{5} \times \frac{12}{13}-\sqrt{1-\left(\frac{3}{5}\right)^{2}} \cdot \sqrt{1-\left(\frac{12}{13}\right)^{2}}\right)+\cos ^{-1} \frac{63}{65} $
$=\cos ^{-1}\left(\frac{36}{65}-\sqrt{\frac{5^{2}-3^{2}}{5^{2}}} \cdot \sqrt{\frac{13^{2}-12^{2}}{13^{2}}}\right)+\cos ^{-1} \frac{63}{65} $
$=\cos ^{-1}\left(\frac{36}{65}-\frac{\sqrt{25-9}}{5} \cdot \frac{\sqrt{169-144}}{13}\right)+\cos ^{-1} \frac{63}{65}$
$=\cos ^{-1}\left(\frac{36}{65}-\frac{4}{5} \times \frac{5}{13}\right)+\cos ^{-1} \frac{63}{65}$
$=\cos ^{-1}\left(\frac{36}{65}-\frac{20}{65}\right)+\cos ^{-1} \frac{63}{65}$
$=\cos ^{-1} \frac{16}{65}+\cos ^{-1} \frac{63}{65}$
$=\cos ^{-1}\left(\frac{16}{65} \times \frac{63}{65}-\sqrt{1-\left(\frac{16}{65}\right)^{2}} \cdot \sqrt{1-\left(\frac{63}{65}\right)^{2}}\right)$
$=\cos ^{-1}\left(\frac{16}{65} \times \frac{63}{65}-\sqrt{\frac{65^{2}-16^{2}}{652}} \cdot \frac{65^{2}-63^{2}}{65^{2}}\right)$
$=\cos ^{-1}\left(\frac{1008}{65 \times 65}-\frac{\sqrt{(65+16)(65-16)}}{65} \cdot \frac{\sqrt{(65+63)(65-63)}}{65}\right)$
$=\cos ^{-1}\left(\frac{1008}{65 \times 65}-\frac{\sqrt{81 \times 49}}{65} \cdot \frac{\sqrt{128 \times 2}}{65}\right)$
$=\cos ^{-1}\left(\frac{1008}{65 \times 65}-\frac{7 \times 9}{65} \times \frac{16}{65}\right)$
$=\cos ^{-1}\left(\frac{1008}{65 \times 65}-\frac{1008}{65 \times 65}\right)$
$=\cos ^{-1}(0)$
$=\cos ^{-1} \cos \frac{\pi}{2}$
Sol :
L.H.S
$=\sin ^{-1} \frac{8}{17}+\sin ^{-1} \frac{3}{5} $
$=\sin ^{-1}\left(\frac{8}{17} \cdot \sqrt{1-\left(\frac{3}{5}\right)^{2}}+\frac{3}{5} \cdot \sqrt{1-\left(\frac{8}{17}\right)^{2}}\right)$
$=\sin ^{-1}\left(\frac{8}{17} \cdot \sqrt{\frac{5^{2}-3^{2}}{5^{2}}}+\frac{3}{5} \cdot\sqrt{\frac{17^{2}-8^{2}}{17^{2}}}\right) $
$=\sin ^{-1}\left(\frac{8}{17} \times \frac{4}{5}+\frac{3}{5} \times \frac{15}{17}\right)$
$=\sin ^{-1}\left(\frac{32}{85}+\frac{45}{85}\right)$
Let $=sin^{-1} \dfrac{77}{85}=\theta$ then $sin\theta= \dfrac{77}{85}$
$tan \theta=\dfrac{77}{36}$
∴$\theta=tan^{-1}\dfrac{77}{36}$
(iv) $\cos ^{-1} \frac{12}{13}+\sin ^{-1} \frac{5}{13}=\sin ^{-1} \frac{56}{65}$ (Not correct)
(v) $\cos ^{-1} \frac{4}{5}+\cos ^{-1} \frac{12}{13}=\cos ^{-1} \frac{33}{65}$
Sol :
L.H.S$=\cos ^{-1} \frac{4}{5}+\cos ^{-1} \frac{12}{13}$
$=\cos ^{-1}\left(\frac{4}{5} \times \frac{12}{13}-\sqrt{1-\frac{16}{25}} \cdot \sqrt{1-\frac{144}{169}}\right) $
$=\cos ^{-1}\left(\frac{48}{65}-\frac{3}{5} \times \frac{5}{13}\right)$
$=\cos ^{-1}\left(\frac{48}{65}-\frac{15}{65}\right) $
$=\cos ^{-1} \frac{33}{65}$=R.H.S proved
(vi) $\sin ^{-1} \frac{3}{5}-\sin ^{-1} \frac{8}{17}=\cos ^{-1} \frac{84}{85}$
Sol:
L.H.S $=\sin ^{-1} \frac{3}{5}-\sin ^{-1} \frac{8}{17}$
$=\sin ^{-1}\left(\frac{3}{5} \cdot \sqrt{1-\left(\frac{8}{17}\right)^{2}}-\frac{8}{17} \sqrt{1-\left(\frac{3}{5}\right)^{2}}\right)$
$=\sin ^{-1}\left(\frac{3}{5} \times \sqrt{\frac{17^{2}-8^{2}}{17^{2}}}-\frac{8}{17} \sqrt{\frac{5^{2}-3^{2}}{5^{2}}}\right)$
$=\sin ^{-1}\left(\frac{3}{5} \times \frac{\sqrt{289-64}}{17}-\frac{8}{17} \frac{\sqrt{25-9}}{5}\right)$
$=\sin ^{-1}\left(\frac{3}{5} \times \frac{15}{17}-\frac{8}{17} \times \frac{4}{5}\right)$
$=\sin ^{-1}\left(\frac{45}{85}-\frac{32}{85}\right)$
$=\sin ^{-1} \frac{13}{85}=\theta$ (Let)
then $\sin \theta=\frac{13}{89} \quad \therefore \cos \theta=\frac{84}{85}$
$\therefore \theta=\cos ^{-1} \frac{84}{85}$
$\sin ^{-1} \frac{13}{85}=\theta=\cos ^{-1} \frac{84}{85}$=R.H.S Proved
(vii) $\sin ^{-1} \frac{5}{13}+\cos \frac{1}{5}=\tan ^{-1} \frac{63}{16}$
Sol :
L.H.S $=\sin ^{-1} \frac{5}{13}+\cos ^{-1} \frac{3}{5}$
$=\sin ^{-1} \frac{5}{13}+\sin ^{-1} \frac{4}{5}$
=$sin^{-1}\left(\dfrac{5}{13}.\sqrt{1-\left(\dfrac{4}{5}\right)^2}+\dfrac{4}{5}\sqrt{1-\left(\dfrac{5}{13}\right)^2}\right)$
$=\sin ^{-1}\left(\frac{5}{13} \cdot \frac{\sqrt{5^{2}-4^{2}}}{5}+\frac{4}{5} \cdot \frac{\sqrt{13^{2}-5^{2}}}{13}\right)$
$=\sin ^{-1}\left(\frac{5}{13} \cdot \frac{3}{5}+\frac{4}{5} \cdot \frac{12}{13}\right)$
$=\sin ^{-1}\left(\frac{15}{65}+\frac{48}{65}\right)$
∴$\tan \theta=\frac{63}{16} \Rightarrow \theta=\tan ^{-1} \frac{63}{16}$
∴$\sin ^{-1} \frac{63}{65}=\theta=\tan ^{-1} \frac{63}{16}$=R.H.S proved
(viii) $\sin ^{-1} \frac{12}{13}+\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{63}{16}=\pi$
Sol :
LH.S $=\sin ^{-1} \frac{12}{13}+\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{63}{16}$
$=\cos ^{-1} \frac{5}{13}+\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{63}{16}$
$=\cos ^{-1}\left(\frac{5}{13} \times \frac{4}{5}-\sqrt{1-\left(\frac{5}{19}\right)^{2}} \cdot \sqrt{1-\left(\frac{4}{5}\right)^{2}}\right)+\tan ^{-1} \frac{63}{16}$
$=\cos ^{-1}\left(\frac{20}{65}-\sqrt{\frac{13^{2}-5^{2}}{13^{2}}} \cdot \sqrt{\frac{5^{2}-4^{2}}{5^2}}\right)+\tan ^{-1} \frac{63}{16}$
$=\cos ^{-1}\left(\frac{20}{65}-\frac{\sqrt{169-25}}{13} \cdot \frac{\sqrt{25-16}}{5}\right)+\tan ^{-1} \frac{63}{16}$
$=\cos ^{-1}\left(\frac{20}{65}-\frac{36}{65}\right)+\tan ^{-1} \frac{63}{16}$
$=\cos ^{-1}\left(-\frac{16}{65}\right)+\tan ^{-1} \frac{63}{16}$
$=\pi-\cos ^{-1} \frac{16}{65}+\tan ^{-1} \frac{63}{16}$
$=\pi-\tan ^{-1} \frac{63}{16}+\tan ^{-1} \frac{63}{16}$
$=\pi$ =R.H.S proved
(ix) $2 \sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{24}{7}$
Sol :
l.H.S$=2 \sin ^{-1} \frac{3}{5}$
$=\sin ^{-1}\left(2 \cdot\dfrac{ 3}{5} \cdot \sqrt{1-\left(\dfrac{3}{5}\right)^{2}}\right)$
[∵ 2sin-1x=sin-1(2x.√1-x2)]
$=sin^{-1} \left(\dfrac{6}{5}.\sqrt{\dfrac{5^2-3^2}{5^2}}\right)$
$=sin^{-1} \left(\dfrac{6}{5}\times \dfrac{4}{5}\right)$
$=sin^{-1}\left(\dfrac{24}{25}\right)$
$=tan^{-1} \dfrac{24}{7}$=R.H.S proved
(x) $\dfrac{9\pi}{8}-\dfrac{9}{4}sin^{-1}\dfrac{1}{3}=\dfrac{9}{4}sin^{-1}\dfrac{2\sqrt{2}}{3}$
Sol :
$\frac{9 \pi}{8}=\frac{9}{4} \sin ^{-1} \frac{2 \sqrt{2}}{3}+\frac{9}{4} \sin ^{-1} \frac{1}{3}$
Now , R.H.S=$=\frac{9}{4}\left(\sin ^{-1} \frac{2 \sqrt{2}}{3}+\sin ^{-1} \frac{1}{3}\right)$
$=\dfrac{9}{4}\left[\sin ^{-1}\left(\dfrac{2 \sqrt{2}}{3}\sqrt{1-\dfrac{1}{9}}+\dfrac{1}{3} \cdot \sqrt{1-\dfrac{8}{9}}\right)\right]$
$=\frac{9}{4}\left[\sin ^{-1}\left(\frac{2 \sqrt{2}}{3} \cdot \frac{\sqrt{9-1}}{3}+\frac{1}{3} \cdot \frac{\sqrt{9-8}}{3}\right)\right]$
$=\frac{9}{4}\left[\sin ^{-1}\left(\frac{2 \sqrt{2}}{3} \cdot \frac{2 \sqrt{2}}{3}+\frac{1}{3} \cdot \frac{1}{3}\right)\right]$
$=\frac{9}{4}\left[\sin ^{-1}\left(\frac{8}{9}+\frac{1}{9}\right)\right]$
$=\frac{9}{4}\left[\sin ^{-1}\left(\frac{9}{9}\right)\right]$
=$\dfrac{9}{4}sin^{-1} sin\dfrac{\pi}{2}$
=$\dfrac{9}{4}\times \dfrac{\pi}{2}=\dfrac{9\pi}{8}$=L.H.S Proved
(xii) $\sin ^{-1} x+\sin ^{-1} y=\cos ^{-1}(\sqrt{1-x^{2}} \cdot \sqrt{1-y^{2}}-x y),$ when
$x \in[0,1], y \in[0,1]$
Sol :
L.H.S=sin-1x+sin-1y
=$cos^{-1}\sqrt{1-x^2}+cos^{-1}\sqrt{1-y^2}$
=$cos^{-1}\left(\sqrt{1-x^{2}} \cdot \sqrt{1-y^{2}}-\sqrt{1-(\sqrt{1-x^{2}})^{2}} \cdot \sqrt{1-(\sqrt{1-y^{2}})^{2}}\right)$
=$\cos ^{-1}(\sqrt{1-x^{2}} \cdot \sqrt{1-y^{2}}-\sqrt{1-\left(1-x^{2}\right)} \cdot \sqrt{1-\left(1-y^{2}\right)})$
=$\cos ^{-1}(\sqrt{1-x^{2}} \cdot \sqrt{1-y^{2}}-\sqrt{1-1+x^{2}} \cdot \sqrt{1-1+y^{2}})$
=$\cos ^{-1}(\sqrt{1-x^{2}} \cdot \sqrt{1-y^{2}}-\sqrt{x^{2}} \cdot \sqrt{y^{2}})$
=$\cos ^{-1}(\sqrt{1-x^{2}} \cdot \sqrt{1-y^{2}}-x y)$=R.H.S Proved
Question 9
Prove that(i) $4\left(sin^{-1}\dfrac{1}{\sqrt{10}+cos^{-1}\dfrac{2}{\sqrt{5}}}\right)=\pi$
Sol :
L.H.S=$=4\left(\sin ^{-1} \frac{1}{\sqrt{10}}+\cos ^{-1} \frac{2}{\sqrt{5}}\right)$
$=4\left(\cos ^{-1} \frac{3}{\sqrt{10}}+\cos ^{-1} \frac{2}{\sqrt{5}}\right)$
$=4\left[\cos ^{-1}\left(\frac{3}{\sqrt{10}} \times \frac{2}{\sqrt{5}}-\sqrt{1-\frac{9}{10}} \cdot \sqrt{1-\frac{4}{5}}\right)\right]$
$=4\left[\cos ^{-1}\left(\frac{6}{\sqrt{50}}-\sqrt{\frac{10-9}{10}} \cdot \sqrt{\frac{5-4}{5}}\right)\right]$
$=4\left[\cos ^{-1}\left(\frac{6}{\sqrt{50}}-\frac{1}{110} \cdot \frac{1}{\sqrt{5}}\right)\right]$
$=4\cos ^{-1}\left(\frac{6}{\sqrt{50}}-\frac{1}{\sqrt{50}}\right)$
$=scos^{-1}\dfrac{6-1}{\sqrt{50}}$
$=4cos^{-1}\dfrac{5}{\sqrt{25}\times \sqrt{2}}$
$=4cos^{-1}\dfrac{5}{5\times \sqrt{2}}$
$=4cos^{-1} \dfrac{1}{\sqrt{2}}=4cos^{-1}cos\dfrac{\pi}{4}$
$4 \times \dfrac{\pi}{4}=\pi$=R.H.S Proved
(ii) cos(2 sin-1x)=1-2x2
Sol :
L.H.S=cos(2 sin-1x)
Let 2 sin-1x =θ
$\sin ^{-1}(2 x \cdot \sqrt{1-x^{2}})=\theta$
$\sin \theta=2 x \cdot \sqrt{1-x^{2}}$
∴$\cos \theta=\sqrt{1-\sin ^{2} \theta}=\sqrt{1-4 x^{2}\left(1-x^{2}\right)}$
$\cos \theta=\sqrt{1-4 x^{2}+4 x^{4}}=\sqrt{4 x^{4}+1-4 x^{2}}$
$\cos \theta=\sqrt{\left(1-2 x^{2}\right)^{2}}=\left(1-2 x^{2}\right)$
θ=cos-1(1-2x2)
∴ 2sin-1x=θ=cos-1(1-2x2)
Now L.H.S=cos(2sin-1x)
=1-2x2=R.H.S Proved
Sol :
L.H.S=cos(sec-1x+cosec-1x)
=$cos \dfrac{\pi}{2}=0$=R.H.S Proved
(iv) $\frac{1}{2} \cos ^{-1} x=\sin ^{-1} \sqrt{\frac{1-x}{2}}=\cos ^{-1} \frac{\sqrt{1+x}}{2}=\tan ^{-1} \frac{\sqrt{1-x^{2}}}{1+x}$
Sol :
L.H.S=$\dfrac{1}{2}cos^{-1}x=\theta$ then $cos^{-1}x=2\theta$
⇒x=cos2θ
⇒x=1-2sin2θ
⇒2sin2θ=1-x
⇒$sin2\theta=\dfrac{1-x}{2}$
⇒$sin\theta =\sqrt{\dfrac{1-x}{2}}$
∴$\dfrac{1}{2}cos^{-1}x=\theta=sin\sqrt{\dfrac{1-x}{2}}$=R.H.S Proved
again , L.H.S$=sin^{-1}\sqrt{\dfrac{1-x}{2}}=\theta$ then $sin\theta =\sqrt{\dfrac{1-x}{2}}$
∴$\cos \theta=\frac{1-\sin ^{2} \theta}{2}=\sqrt{1-\frac{1-x}{2}}=\sqrt{\frac{2-1+x}{2}}$
$\cos \theta=\sqrt{\frac{1+x}{2}}$
∴$\theta=cos^{-1}\sqrt{\dfrac{1+x}{2}}$
Now L.H.S=$\sin ^{-1} \sqrt{\frac{1-x}{2}}=\theta=\cos ^{-1} \sqrt{\frac{1+x}{2}}$=R.H.S Proved
Question 10
Prove that
(i) $\sin ^{-1} x+\cos ^{-1} y=\tan ^{-1}\left(\frac{x y+\sqrt{\left(1-x^{2}\right)\left(1-y^{2}\right)}}{y \sqrt{\left(1-x^{2}\right)}-x \sqrt{\left(1-y^{2}\right)})}\right.$
Sol :
L.H.S$=\sin ^{-1} x+\cos ^{-1} y $
$=\cos ^{-1} \sqrt{1-x^{2}}+\cos ^{-1} y $
$=\cos ^{-1}(\sqrt{1-x^{2}} \cdot y-\sqrt{1-\left(1-x^{2}\right)} \cdot \sqrt{1-y^{2}}) $
$=\cos ^{-1}(y \cdot \sqrt{1-x^{2}}-\sqrt{x^{2}} \cdot \sqrt{1-y^{2}})$
$\sin \theta=\sqrt{1-\cos ^{2} \theta}$
$\sin \theta=\sqrt{1-(y \sqrt{1-x^{2}}-x \sqrt{1-y^{2}})^{2}}$
$\sin \theta=\sqrt{1-\left[y^{2}\left(1-x^{2}\right)+x^{2}\left(1-y^{2}\right)-2 x y \sqrt{1-x^{2}}\right. \sqrt{1-y^2}}$
$\sin \theta=\sqrt{1-\left(y^{2}-x^{2} y^{2}+x^{2}-x^{2} y^{2}-2 x y \sqrt{1-x^{2}} \cdot \sqrt{1-y^{2}}\right)}$
$\sin \theta=\sqrt{1-x^{2}-y^{2}+2 x^{2} y^{2}+2 x y \sqrt{1-x^{2}} \cdot \sqrt{1-y^{2}}}$
$\sin \theta=x y+\sqrt{\left(1-x^{2}\right)} \cdot \sqrt{\left(1-y^{2}\right)}$
∴$\tan \theta=\frac{\sin \theta}{\cos \theta}=\frac{x y+\sqrt{\left(1-x^{2}\right)} \cdot \sqrt{\left(1-y^{2}\right)}}{y \cdot \sqrt{\left(1-x^{2}\right)}-x \sqrt{\left(1-y^{2}\right)}}$
Now, $\cos ^{-1}(y \sqrt{1-x^{2}}-x \sqrt{1-y^{2}})=\theta$
∴$\sin ^{-1} x+\cos ^{-1} y=\tan ^{-1}\left(\frac{x y+\sqrt{\left(1-x^{2}\right)} \cdot \sqrt{\left(1-y^{2}\right)}}{y \cdot \sqrt{\left(1-x^{2}\right)-x \sqrt{1-y^{2}}}}\right)$
Proved
Sol :
Given $\tan ^{-1} x+\tan ^{-1} y$ $=\frac{1}{2} \sin ^{-1} \frac{2(x+y)(1-x y)}{\left(1+x^{2}\right)\left(1+y^{2}\right)}$
$=2\left(\tan ^{-1} x+\tan ^{-1} y\right)=\sin ^{-1} \frac{2(x+y)(1-x y)}{\left(1+x^{2}\right)\left(1+y^{2}\right)}$
Now, L.H.S $=2\left(\tan ^{-1} x+\tan ^{-1} y\right)$
$=2 \tan ^{-1}\left(\frac{x+y}{1-x y}\right)$
$\therefore 2tan^{-1}x=sin^{-1}\dfrac{2x}{1+x^2}$
=$\sin ^{-1}\dfrac{\frac{ 2(x+y)}{(1-x y)} \times(1-x y)^{2}}{(1-xy)^2+(x+y)^2}$
=$\sin ^{-1} \dfrac{2(x+y)(1-x y)}{1+x^{2} y^{2}-2 x y+x^{2}+y^{2}+2 x y}$
=$\sin ^{-1} \dfrac{2(x+y)(1-x y)}{1+x^{2}+x^{2} y^{2}+y^{2}}$
=$\sin ^{-1} \dfrac{2(x+y)(1-x y)}{\left(1+x^{2}\right)+y^{2}\left(1+x^{2}\right)}$
=R.H.S Proved
Question 11
Prove that:(i) 2tan-1(cosec.tan-1x-tan.cot-1x)=tan-1x
Sol :
L.H.S=2tan-1(cosec.tan-1x-tan.cot-1x)
$=2 \tan ^{-1}\left(\operatorname{cosec} \tan ^{-1} x-\tan \cdot \tan ^{-1} \frac{1}{x}\right)$
$=2 \tan ^{-1}\left(\operatorname{cosec} \cdot \operatorname{cosec}^{-1} \frac{\sqrt{1+x^{2}}}{x}-\tan \tan \frac{-1}{x}\right)$
=$2 \tan ^{-1}\left(\frac{\sqrt{1+x^{2}}}{x}-\frac{1}{x}\right)$
$=2 \tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)$
$=\tan ^{-1} \frac{2\left(\frac{\sqrt{1+x^{2}-1}}{x}\right)}{1-\left(\frac{\sqrt{1+x^{2}-1}}{x}\right)^{2}}$
$=\tan ^{-1} \frac{2(\sqrt{1+x^{2}}-1)}{\frac{x^{2}-(\sqrt{1+x^{2}-1})^{2}}{x^{2}}}$
=$\tan ^{-1} \frac{2(\sqrt{1+x^{2}}-1) x}{x^{2}-\left(1+x^{2}+1-2 \sqrt{1+x^{2}}\right)}$
$=\tan ^{-1} \frac{2(\sqrt{1+x^{2}}-1) \cdot x}{x^{2}-1-x^{2}-1+2 \sqrt{1+x^{2}}}$
$=\tan ^{-1} \frac{2(\sqrt{1+x^{2}-1}) \cdot x}{2 \sqrt{1+x^{2}-2}}$
$=\tan ^{-1} \frac{2(\sqrt{1+x^{2}}-1) \cdot x}{2(\sqrt{1+x^{2}-1})}$
(ii) $\cos. \tan ^{-1} .\sin .\cot ^{-1} x=\sqrt{\frac{x^{2}+1}{x^{2}+2}}$
Sol :
L.H.S $=\cos. \tan ^{-1} .\sin. \cot ^{-1} x$
$=\cos. \tan ^{-1}. \sin. \sin ^{-1} \frac{1}{\sqrt{1+x^{2}}}$ $\left(\because cot^{-1}=sin^{-1}\dfrac{1}{\sqrt{1+x^2}}\right)$
=$cos.cos^{-1}\dfrac{\sqrt{x^2+1}}{\sqrt{x^2+2}}$
=$\dfrac{\sqrt{x^2+1}}{\sqrt{x^2+2}}$=R.H.S proved
Explain in detail sir jii
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