The CBSE Board students widely use the RD Sharma Solution books to clear their doubts in their homework and score good marks in their public examinations. However, every student cannot depend on a tutor to help them clear their doubts 24 x 7. Significantly, the students who find it difficult to cope with the Inverse Trigonometry portions can use the RD Sharma Class 12th exercise 3.8 books.
RD Sharma Class 12 Solutions Chapter3 Inverse Trigonometric Functions-Other Exercise
Inverse Trigonometric Functions Excercise:3.8
Inverse trigromtery functions exercise 3.8 question 1 (I)
Hint:As we know that the value of
$\sin \left(\sin ^{-1} x\right) \text { is } x \text { for } x \in[-1,1]$Given:We have,
$\sin \left(\sin ^{-1} \frac{7}{25}\right)$Solution:So here in the expression already present the form of
$\sin \left(\sin ^{-1} x\right)$$\sin \left(\sin ^{-1} \frac{7}{25}\right)=\frac{7}{25} \mid$Inverse trigromtery functions exercise 3.8 question 1 (II)
Answer:
$\frac{12}{13}$Hint:As we know that the value of
$\sin \left(\sin ^{-1} x\right) \text { is } x \text { for } x \in[-1,1]$Given:We have,
$\sin \left(\cos ^{-1} \frac{5}{13}\right)$Solution:Here in the place of
$\sin ^{-1} x$ there is
$\cos ^{-1} x$.
So we convert
$\cos ^{-1} x$ into
$\sin ^{-1} x$.
Let’s suppose that
$\begin{aligned} &\cos ^{-1} \frac{5}{13}=\alpha \\ &\cos \alpha=\frac{5}{13} \\ &\sin \left(\cos ^{-1} \frac{5}{13}\right)=\sin \alpha \end{aligned}$Let, in
$\Delta ABC$, angle
$CAB = \alpha$ and right angle at B
$\cos \alpha=\frac{5}{13}=\frac{B}{H}$
$\begin{aligned} &(A C)^{2}=(A B)^{2}+(B C)^{2} \\ &(13)^{2}=5^{2}+(B C)^{2} \\ &(B C)^{2}=169-25 \end{aligned}$$BC= \pm 12$ [we will ignore the -ve sign because BC is a length and it can’t be -ve.]
$BC= 12$$\sin =\frac{12}{13}$So, we required
$\alpha$ and that is
$\frac{12}{13}$$\sin \left(\cos ^{-1} \frac{5}{13}\right)=\sin \alpha=\frac{12}{13}$Inverse trigromtery functions exercise 3.8 question 1 (III)
Answer:
$\frac{24}{25}$Hint:As we know that the value of
$\left(\sin ^{-1} x\right) \text { is } x \text { for } x \in[-1,1]$Given:We have
$\sin \left(\tan ^{-1}\left(\frac{24}{7}\right)\right)$Solution:Here in the place of
$\sin ^{-1} x$we have $\tan ^{-1} x$.So, we convert
$\tan ^{-1} x$ to $\sin ^{-1} x$.
Let’s suppose that,
$\begin{aligned} &\tan ^{-1}\left(\frac{24}{7}\right)=\alpha \\ &\tan \alpha=\frac{24}{7} \end{aligned}$Let, in
$\Delta ABC$, angle
$CAB = \alpha$ and right angle at B.
$\begin{aligned} \tan \alpha &=\frac{P}{B}=\frac{24}{7} \\ (A C)^{2} &=(A B)^{2}+(B C)^{2} \\ &=7^{2}+(24)^{2} \\ &=49+576 \\ &=625 \end{aligned}$$= +25$ [we will ignore the -ve sign because AC is a length and it can’t be -ve.]
$AC= 25$$\begin{aligned} &\sin \alpha=\frac{24}{25} \\ &\sin \left(\tan ^{-1}\left(\frac{24}{7}\right)\right)=\sin \alpha=\frac{24}{25} \end{aligned}$ [since,
$\tan ^{-1}\left(\frac{24}{7}\right)=\alpha$]
Inverse trigonometric functions exercise 3.8 question 1 (iv)
Answer:
$\frac{15}{17}$Hint:As we know that the value of
$\left(\sin ^{-1} x\right) \text { is } x \text { for } x \in[-1,1]$Given:We have
$\sin \left(\sec ^{-1}\left(\frac{17}{8}\right)\right)$Solution:So here in place of
$\sin ^{-1} x$ there is
$\sec ^{-1} x$.
So, we convert
$\sec ^{-1} x$ into
$\sin ^{-1} x$Let’s we suppose that,
$\begin{aligned} &\sec ^{-1} \frac{17}{8}=\alpha \\ &\sec \alpha=\frac{17}{8} \end{aligned}$Let, in
$\Delta ABC$, angle
$CAB = \alpha$ and right angle at B.
$\begin{aligned} &\sec \alpha=\frac{H}{B}=\frac{17}{8} \\ &(A C)^{2}=(A B)^{2}+(B C)^{2} \\ &(17)^{2}=8^{2}+(B C)^{2} \\ &(B C)^{2}=289-64 \end{aligned}$$BC =\pm 15$ [we will ignore the -ve sign because BC is a length and it can’t be -ve]
$BC = 15$$\begin{aligned} &\sin \alpha=\frac{P}{H}=\frac{15}{17} \\ &\sin \left(\sec ^{-1}\left(\frac{17}{8}\right)\right)=\frac{15}{17} \end{aligned}$Inverse trigonometric functions exercise 3.8 question 1 (v)
Answer:
$\frac{5}{4}$Hint:As we know that the value of
$\cos e c\left(\cos e c^{-1} x\right) \text { is } x \text { for } x \in(-\infty,-1) \cup(1, \infty)$Given:We have
$\operatorname{cosec}\left(\cos ^{-1} \frac{3}{5}\right)$Solution:
In place of
$\cos e c^{-1} x$ here is
$\cos ^{-1} x$So, we convert
$\cos ^{-1} x$ into
$\cos e c^{-1} x$.
Let’s suppose that,
$\begin{aligned} &\cos ^{-1} \frac{3}{5}=\alpha \\ &\cos \alpha=\frac{3}{5}=\frac{B}{H} \end{aligned}$Let, in
$\Delta ABC$, angle
$CAB = \alpha$ and right angle at B.
$\begin{aligned} &(A C)^{2}=(A B)^{2}+(B C)^{2} \\ &5^{2}=3^{2}+(B C)^{2} \\ &25-9=(B C)^{2} \end{aligned}$$BC =\pm 4$ [we will ignore the -ve sign because BC is a length and it can’t be -ve]
$BC =4$$\begin{aligned} &\cos e c \alpha=\frac{H}{P}=\frac{5}{4} \\ &\cos e c\left(\cos ^{-1} \frac{3}{5}\right)=\frac{5}{4} \end{aligned}$Inverse trigonometric functions exercise 3.8 question 1 (vi)
Answer:
$\frac{13}{5}$Hint:As we know that the range of
$\sec \left(\sec ^{-1} x\right) \text { is } x \text { for } x \in(-\infty,-1) \cup(1, \infty)$Given:We have
$\sec \left(\sin ^{-1}\left(\frac{12}{13}\right)\right)$Solution:In place of
$\sec ^{-1} x$ there is
$\sin ^{-1} x$So, we convert
$\sin ^{-1} x$ into
$\sec ^{-1} x$ .
Let’s suppose that,
$\begin{aligned} &\sin ^{-1} \frac{12}{13}=\alpha \\ &\sin \alpha=\frac{12}{13} \end{aligned}$Let, in
$\Delta ABC$, angle
$CAB = \alpha$ and right angle at B.
$\begin{aligned} &(A C)^{2}=(A B)^{2}+(B C)^{2} \\ &(13)^{2}=(AB)^{2}+(12)^{2} \\ &(AB)^2=25 \end{aligned}$$AB = \pm 5$ [we will ignore the -ve sign because AB is a length and it can’t be -ve]
$AB = 5$$\begin{aligned} &\sec \alpha=\frac{13}{5} \\ &\sec \left(\sin ^{-1}\left(\frac{12}{13}\right)\right)=\sec \alpha=\frac{13}{5} \\ &\sec \left(\sin ^{-1}\left(\frac{12}{13}\right)\right)=\frac{13}{5} \end{aligned}$Inverse Trigonometric functions exercise 3.8 question 1 (vii)
Answer:
$\frac{15}{8}$Hint:As we know that the value of
$\tan \left(\tan ^{-1} x\right) \text { is } x \text { for } x \in \mathbb{R} \text { ( real number) }$Given:We have
$\tan \left(\cos ^{-1}\left(\frac{8}{17}\right)\right)$Solution:In place of
$\tan ^{-1} x$ here is
$\cos ^{-1} x$So, we convert
$\cos ^{-1} x$ into
$\tan ^{-1} x$.
Let’s suppose that,
$\begin{aligned} &\cos ^{-1} \left (\frac{8}{17} \right )=\alpha \\ &\cos \alpha=\frac{8}{17}=\frac{B}{H} \end{aligned}$Let, in
$\Delta ABC$, angle
$CAB = \alpha$ and right angle at B.
$\begin{aligned} &\\ &(A C)^{2}=(A B)^{2}+(B C)^{2} \\ &(17)^{2}=8^{2}+(B C)^{2} \\ &(B C)^{2}=289-64 \\ & (BC)^2= 25\end{aligned}$$BC=\pm 15$ [we will ignore the -ve sign because BC is a length and it can’t be -ve]
$BC= 15$$\begin{aligned} &\tan \left(\cos ^{-1}\left(\frac{8}{17}\right)\right)=\tan \alpha \\ &\tan \alpha=\frac{15}{8} \\ &\tan \left(\cos ^{-1}\left(\frac{8}{17}\right)\right)=\frac{15}{8} \end{aligned}$Inverse Trigonometric functions exercise 3.8 question 1 (viii)
Answer:
$\frac{3}{4}$Hint:As we know that the range of
$\cot \left(\cot ^{-1} x\right) \text { is } x \text { for } x \in \mathbb{R} \text { ( real number) }$Given:$\cot \left(\cos ^{-1}\left(\frac{3}{5}\right)\right)$Solution:In place of
$\cot ^{-1} x$ here is
$\cos ^{-1} x$So, we convert
$\cos ^{-1} x$ into
$\cot ^{-1} x$.
Let’s suppose that,
$\begin{aligned} &\cos ^{-1} \left (\frac{3}{5} \right )=\alpha \\ &\cos \alpha=\frac{3}{5}=\frac{B}{H} \end{aligned}$Let, in
$\Delta ABC$, angle
$CAB = \alpha$ and right angle at B.
$\begin{aligned} &\\ &(A C)^{2}=(A B)^{2}+(B C)^{2} \\ &(5)^{2}=3^{2}+(B C)^{2} \\ &25=9 + (BC)^2 \\ & (BC)^2= 16\end{aligned}$$BC=\pm 4$ [we will ignore the -ve sign because BC is a length and it can’t be -ve]
$BC= 4$$\begin{aligned} &\cot \left(\cos ^{-1}\left(\frac{3}{5}\right)\right)=\cot c \\ &\cot \alpha=\frac{3}{4} \\ &\cot \left(\cos ^{-1}\left(\frac{3}{5}\right)\right)=\frac{3}{4} \end{aligned}$
Inverse Trigonometric functions exercise 3.8 question 1 (ix)
Answer:
$\frac{7}{25}$Hint:As we know that the range of
$\left(\cos ^{-1} x\right) \text { is } x \text { for } x \in[-1,1]$Given:We have
$\cos \left(\tan ^{-1}\left(\frac{24}{7}\right)\right)$Solution:In place of
$\cos ^{-1} x$ here is
$\tan ^{-1} x$So, we convert
$\tan ^{-1} x$ into
$\cos ^{-1} x$ .
Let’s suppose that,
$\begin{aligned} &\tan ^{-1} \left (\frac{24}{7} \right )=\alpha \\ &\tan \alpha=\frac{24}{7} \end{aligned}$Let, in
$\Delta ABC$, angle
$CAB = \alpha$ and right angle at B.
$\begin{aligned} &\\ &(A C)^{2}=(A B)^{2}+(B C)^{2} \\ &(AC)^{2}=7^{2}+(24)^{2} \\ &(A C)^{2}=625 \end{aligned}$$AC=\pm 25$ [we will ignore the -ve sign because AB is a length and it can’t be -ve]
$AC=25$$\begin{aligned} &\cos \left(\tan ^{-1}\left(\frac{24}{7}\right)\right)=\cos \alpha \\ &\cos \alpha=\frac{7}{25} \\ &\cos \left(\tan ^{-1}\left(\frac{24}{7}\right)\right)=\frac{7}{25} \end{aligned}$Inverse Trigonometric functions exercise 3.8 question 2 (i)
Hint:We convert in the form of
$\tan (\alpha + \beta)$ so we know the formula of it.
Given:We have to prove that
$\tan \left(\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{2}{3}\right)=\frac{17}{6}$Solution:LHS=
$\tan \left(\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{2}{3}\right)$Let suppose that
$\cos ^{-1}\frac{4}{5}=\alpha$ and
$\tan ^{-1}\frac{2}{3}=\beta$,
$\begin{aligned} &\cos \alpha=\frac{4}{5} \text { and } \tan \beta=\frac{2}{3} \\ &\tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta} \end{aligned}$But we have not
$\tan \alpha$. We have only
$\tan \beta$. So, we convert
$\cos \alpha$ into
$\tan \alpha$.
$\cos \alpha =\frac{4}{5}=\frac{B}{H}$Let, in
$\Delta ABC$, angle
$CAB = \alpha$ and right angle at B.
$\begin{aligned} &\\ &(A C)^{2}=(A B)^{2}+(B C)^{2} \\ &(5)^{2}=4^{2}+(B C)^{2} \\ &(B C)^{2}=25-16 \\ \end{aligned}$$BC =\pm 9$$BC = 3$$\tan \alpha = \frac{3}{4}$$[\because \tan \alpha = \frac{3}{4} \; and\: \tan \beta = \frac{2}{3} ]$$\begin{gathered} \tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta} \\ =\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4} \times \frac{2}{3}} \\ =\frac{\frac{9+8}{12-6}}{12}=\frac{17}{6}=\mathrm{RHS} \\ \tan \left(\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{2}{3}\right)=\frac{17}{6} \end{gathered}$ [ we take LCM of 3,4]
Hence proved.
Inverse trigonometric functions exercise 3.8 question 2 (ii)
Hint:We convert in the form of
$\cos (\alpha + \beta)$and we know the formula of it.
Given:We have to prove
$\cos \left(\sin ^{-1} \frac{3}{5}+\cot ^{-1} \frac{3}{2}\right)=\frac{6}{5 \sqrt{13}}$Solution: LHS =
$\cos \left(\sin ^{-1} \frac{3}{5}+\cot ^{-1} \frac{3}{2}\right)$Let’s suppose that
$\sin ^{-1} \frac{3}{5}= \alpha$and
$\cot ^{-1} \frac{3}{2}=\beta$.
$\sin \alpha = \frac{3}{5} \; and\; \cot \beta =\frac{3}{2}$$\cos\left ( \alpha +\beta \right )= \cos \alpha \cos \beta -\sin \alpha \sin \beta$Here, we only have
$\sin \alpha$ and
$\cos \beta$.So we have to find
$\cos \alpha$and
$\sin \beta$.
$\sin \alpha =\frac{3}{5}=\frac{B}{H}$Let, in
$\Delta ABC$, angle
$CAB = \alpha$ and right angle at B.
$\begin{aligned} &\\ &(A C)^{2}=(A B)^{2}+(B C)^{2} \\ &(5)^{2}=(AB)^{2}+3^{2} \\ &(AB )^{2}=25-9 \\ \end{aligned}$$AB=\pm 4$$\cos \alpha = \frac{4}{5}$Let, in
$\Delta PQR$, angle
$RPQ = \beta$ and right angle at B.
$\cot \beta =\frac{3}{2}=\frac{B}{P}$
$\begin{aligned} &\\ &(PR)^{2}=(QR)^{2}+(PQ)^{2} \\ &(PR)^{2}=2^{2}+3^{2} \\ &(PR)^{2}=4 + 9 \\ \end{aligned}$$PR =\pm \sqrt{13}$ [we will ignore the -ve sign because PR is a length and it can’t be -ve]
$PR = \sqrt{13}$$\begin{aligned} &P R=\sqrt{13} \\ &\sin \beta=\frac{2}{\sqrt{13}}, \cos \beta=\frac{3}{\sqrt{13}} \\ &\cos (\alpha+\beta)=\cos \alpha \cos \beta-\sin \alpha \sin \beta \\ &=\frac{4}{5} \times \frac{3}{\sqrt{13}}-\frac{2}{\sqrt{13}} \times \frac{3}{5} \\ &=\frac{12}{5 \sqrt{13}}-\frac{6}{5 \sqrt{13}} \\ &\cos (\alpha+\beta)=\frac{6}{5 \sqrt{13}}=\mathrm{RHS} \end{aligned}$Inverse trigonometric functions exercise 3.8 question 2 (iii)
Hint:We convert expression in the form of
$\tan (\alpha + \beta)$ so we know the formula of it.
Given:We have to prove
$\tan \left(\sin ^{-1} \frac{5}{13}+\cos ^{-1} \frac{3}{5}\right)=\frac{63}{16}$Solution:LHS =
$\tan \left(\sin ^{-1} \frac{5}{13}+\cos ^{-1} \frac{3}{5}\right)$$\sin ^{-1} \frac{5}{13}=\alpha \; and \; \cos ^{-1} \frac{3}{5}=\beta$$\sin \alpha =\frac{5}{13} \; and \; \cos \beta = \frac{3}{5}$$\tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}$ …(i)
But we have not
$\tan \alpha$ and
$\tan \beta$. So, we find out first then we put in equation (i)
$\sin \alpha = \frac{5}{13}=\frac{P}H{}$Let, in
$\Delta ABC$, angle
$CAB = \alpha$ and right angle at B.
$\begin{aligned} &\\ &(A C)^{2}=(A B)^{2}+(B C)^{2} \\ &(13)^{2}=AB^{2}+5^{2} \\ &(AB)^{2}=144 \\ \end{aligned}$$AB =\pm 12$ [we will ignore the -ve sign because AB is a length and it can’t be -ve]
$AB =12$$\therefore \tan \alpha = \frac{5}{12}$Let, in
$\Delta PQR$, angle
$RPQ = \beta$ and right angle at B.
$\cot \beta =\frac{3}{5}=\frac{B}{P}$
$\begin{aligned} &\\ &(PR)^{2}=(QR)^{2}+(PQ)^{2} \\ &(5)^{2}=3^{2}+(QR)^{2} \\ &(QR)^{2}=16\\ \end{aligned}$$QR = \pm 4$$QR = 4$$\therefore \tan \beta = \frac{4}{3}$Now,
$\begin{aligned} \tan (\alpha+\beta)=& \frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta} \\ &=\frac{\frac{5}{12}+\frac{4}{3}}{1-\frac{5}{12} \times \frac{4}{3}} \end{aligned}$ [ put,
$\tan \alpha = \frac{5}{12}$ and
$\tan \beta = \frac{4}{3}$]
[ we take LCM of 3 and 12]
$\begin{aligned} &=\frac{\frac{5+16}{12}}{\frac{36-20}{36}} \\ &=\frac{\frac{21}{12}}{\frac{16}{36}} \\ &=\frac{21}{12} \times \frac{36}{16} \\ &=\frac{21}{1} \times \frac{3}{16} \\ \tan (\alpha+\beta) &=\frac{63}{16}=\mathrm{RHS} \end{aligned}$Hence proved.
Inverse trigonometric functions exercise 3.8 question 2 (iv)
Answer:
Hint:We convert in the form of
$\sin (\alpha + \beta)$and we know the formula of it.
Given:We have to prove
$\sin \left(\cos ^{-1} \frac{3}{5}+\sin ^{-1} \frac{5}{13}\right)=\frac{63}{65}$Solution: LHS =
$\sin \left(\cos ^{-1} \frac{3}{5}+\sin ^{-1} \frac{5}{13}\right)$Let’s suppose that
$\cos ^{-1} \frac{3}{5}=\alpha \; and \; \sin ^{-1} \frac{5}{13}=\beta$$\cos \alpha= \frac{3}{5} \; and \; \sin \beta \frac{5}{13}$ and
$\begin{aligned} &\Rightarrow \quad \sin \left(\cos ^{-1} \frac{3}{5}+\sin ^{-1} \frac{5}{13}\right) \\ &\Rightarrow \quad \sin (\alpha+\beta)=\sin \alpha \cos \beta+\cos \alpha \sin \beta \end{aligned}$…(i)
Here we don’t have
$\sin \alpha . \cos \beta$. So, we find out it and then put all the values in equation (i)
Let, in
$\Delta ABC$, angle
$CAB = \alpha$ and right angle at B.
$\cos \alpha = \frac{3}{5}=\frac{B}H{}$
$\begin{aligned} &\\ &(A C)^{2}=(A B)^{2}+(B C)^{2} \\ &(5)^{2}=(3)^{2}+(BC)^{2} \\ &25^{2}=9^2+(BC)^2 \\ & (BC)^2 =16 \end{aligned}$$BC=\pm 4$$BC= 4$$\therefore \sin \alpha =\frac{4}{5}$Let, in
$\Delta PQR$, angle
$RPQ = \beta$ and right angle at B.
$\sin \beta = \frac{5}{13}=\frac{P}H{}$
$\begin{aligned} &\\ &(PR)^{2}=(PQ)^{2}+(QR)^{2} \\ &(13)^{2}=(PQ)^{2} + 5^2\\ &169=(PQ)^2+25\\ & (PQ)^2 =144 \end{aligned}$$PQ=\pm 12$ [we will ignore the -ve sign because PQ is a length and it can’t be -ve]
$PQ= 12$$\therefore \cos \beta =\frac{12}{13}$Now,
$\begin{aligned} \sin (\alpha+\beta) &=\sin \alpha \cos \beta+\cos \alpha \sin \beta \\ &=\frac{4}{5} \times \frac{12}{13}+\frac{3}{5} \times \frac{5}{13} \\ &=\frac{48}{65}+\frac{15}{65} \\ \sin (\alpha+\beta) &=\frac{63}{65}=\mathrm{RHS} \end{aligned}$Hence proved.
Inverse trigromtery functions exercise 3.8 question 3 (i)
Answer:
$x= \pm \sqrt{\frac{35}{36}}$Hint:We know the value of
$\operatorname{cos}\left(\cos ^{-1} x\right) \text { is } x \text { for } x \in[-1,1]$Given:We have to solve
$\cos \left (\sin ^{-1} x \right )= \frac{1}{6}$ to find the value of x.
Solution:Here, we have given
$\sin ^{-1} x$ in place of
$\cos ^{-1} x$. So, we convert
$\sin ^{-1} x$ into
$\cos ^{-1} x$Let’s suppose that,
$\sin ^{-1} x =\alpha$$\sin \alpha =x$We use the identity,
$\begin{aligned} &\sin ^{2} \alpha+\cos ^{2} \alpha=1 \\ &x^{2}+\cos ^{2} \alpha=1 \\ &\cos ^{2} \alpha=1-x^{2} \end{aligned}$$\cos \alpha =\pm \sqrt{1-x^2}$ [But we can eliminate negative value of
$\cos \alpha$ , because in RHS of the question only positive value is given.]
$\cos \alpha =\sqrt{1-x^2}$ …(i)
Now,
$\begin{aligned} &\cos \left(\sin ^{-1} x\right)=\frac{1}{6} \\ &\cos \alpha=\frac{1}{6} \end{aligned}$ .....(ii)
From equation (i) and (ii) we get,
$\sqrt{1-x^2}=\frac{1}{6}$Squaring on both sides,
$\begin{aligned} &\left(\sqrt{1-x^{2}}\right)^{2}=\left(\frac{1}{6}\right)^{2} \\ &1-x^{2}=\frac{1}{36} \\ &x^{2}=1-\frac{1}{36} \\ &x^{2}=\frac{35}{36} \\ &x=\pm \sqrt{\frac{35}{36}} \end{aligned}$Therefore, the answer is
$x= \pm \sqrt{\frac{35}{36}}$ .
Inverse trigromtery functions exercise 3.8 question 3 (ii)
Answer:
$x=\pm \frac{1}{\sqrt{2}}$
Hint:
As we know that the value of$\operatorname{cos}\left(\cos ^{-1} x\right) \text { is } x \text { for } x \in[-1,1]$
Given:
We have to solve $\cos \left(2 \sin ^{-1}(-x)\right)=0$ to find the value of x.
Solution:
So, we convert $2\sin ^{-1}(x)$ into $\cos ^{-1}(x)$
$\cos \left(2 \sin ^{-1}(-x)\right)=\cos \left(-2 \sin ^{-1} x\right)$ [$\because \sin ^{-1}(-x)=-\sin x$]
$=\cos \left(-2 \sin ^{-1} x\right)$ …(i) [ as, $\cos$ is even function, so $\cos(-x)= \cos x$]
Let’s suppose,
$\sin ^{-1} x =\alpha$ …(ii)
$\sin \alpha =x$ …(iii)
Given,
$\cos \left(2 \sin ^{-1}(-x)\right)=0$
$\cos \left(2 \sin ^{-1}(-x)\right)=0$ … [ from eqn(i) ]
$\cos 2 \alpha =0$ …[ using (ii) ]
$1-2 \sin ^2 \alpha =0$ [$\cos 2 \alpha =1-2 \sin ^2 \alpha$ ]
$\begin{aligned} &1-2 x^{2}=0 \\ &2 x^{2}=1 \\ &x^{2}=\frac{1}{2} \\ &x=\pm \frac{1}{\sqrt{2}} \end{aligned}$
Therefore, the answer is $x=\pm \frac{1}{\sqrt{2}}$
The 3rd chapter in mathematics, Inverse Trigonometry for Class 12, is not easy for many students to understand. Especially in the eighth exercise, which is Exercise 3.8, the students must solve the functions as learned from the previous exercises. The concepts given here are regarding solving sine, cosine, tangent, cotangent, secant, and cosecant functions. There are about fifteen questions, including the subparts in this exercise 3.8. If they find it hard, the RD Sharma Class 12 Chapter 3 Exercise 3.8 solutions will lend them a helping hand.
The solutions in this book are created by experts and verified according to the updated syllabus. You can use this guide to complete your homework and make your assignments too. As it follows the NCERT pattern, it is beneficial for the students who pursue their education at CBSE institutions. If you see yourself scoring low marks in the chapter Inverse Trigonometry, use the solutions given in the Class 12 RD Sharma Chapter 3 Exercise 3.8 Solution and understand the concepts. The different ways to solve a single question are given in the material. Soon, you will find yourself crossing your benchmark score.
Looking at the benefits, you attain from this RD Sharma book, you might presume that it might cost a lot. But that is not the fact. The RD Sharma Class 12 Solutions Inverse Trigonometry Ex 3.8 can be downloaded for free from top websites like Career360. All you need is to visit the Careers360 website and search for the solutions of Class 12 Mathematics Chapter 3 for any exercises that you need. You will be able to access the best set of RD Sharma solutions free of cost.
Due to the wide recognition of the solutions provided in the RD Sharma book, there are many chances that you will be asked questions from this book at your public examinations. Many students have benefitted by using the RD Sharma Class 12 Solutions Chapter 3 ex 3.8 to understand the concepts in Inverse Trigonometry deeply. Download your set of RD Sharma solutions book now from the Career360 website.
Chapter-wise RD Sharma Class 12 Solutions