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- Question 5, Exercise 10.1
- quadrant, therefor it lies in 3rd quadrant. Now \begin{align}{{\sec}^{2}}\alpha &=1+{{\tan}^{2}}\alpha\\... is in the 3rd quadrant, value of $\sec$ is –ive \begin{align}\sec \alpha &=-\sqrt{1+{{\tan}^{2}}\alpha }... \Rightarrow \quad \cos\alpha=-\dfrac{4}{5}$$ Now \begin{align}\frac{\sin \alpha }{\cos \alpha }&=\tan \al... ign} Given: $\sec \beta =\dfrac{13}{5}.$ As \begin{align}\cos\beta &= \dfrac{1}{\sec\beta} = \dfrac{
- Question 13, Exercise 10.1
- phi$ and $3=r\sin\varphi$. Squaring and adding \begin{align} &r^2\cos^2 \varphi+r^2\sin^2\varphi = 4^2+... lies &r^2 = 25\\ \implies &r=5 \end{align} Also \begin{align}r\cos\varphi = 4 \implies 5\cos\varphi = 4 ... implies \cos\varphi =\dfrac{4}{5}\end{align} and \begin{align}r\sin\varphi = 3 \implies 5\sin\varphi = 3 ... \dfrac{3}{5}.\end{align} Thus, from (1), we have \begin{align}&4\sin\theta +3\cos\theta \\ &=5 \left(\dfr
- Question 7, Exercise 10.2
- eta =\dfrac{1}{\sec 2\theta }$. ====Solution==== \begin{align}L.H.S&={{\cos }^{4}}\theta -{{\sin }^{4}}\t... }{2}=\dfrac{2}{\sin \theta }$. ====Solution==== \begin{align}L.H.S&=\tan \dfrac{\theta }{2}+co\operatorn... \theta }={{\cot }^{2}}\theta $. ====Solution==== \begin{align}L.H.S&=\dfrac{1+\cos 2\theta }{1+\cos 2\the... eta -\cot 2\theta =tan\theta $. ====Solution==== \begin{align}L.H.S.&=\cos ec2\theta -\cot 2\theta \\ &=
- Question 3, Exercise 10.1
- uadrant and $\cos $ is $+ve$ in first quadrant . \begin{align}\Rightarrow \,\,\,\,\,\cos u&=\sqrt{1-{{\si... quadrant and $\cos $ is $+ve$ in first quadrant. \begin{align}\Rightarrow \,\,\,\,\,\cos v&=\sqrt{1-{{\si... rrow \,\,\,\cos v &=\frac{3}{5}\end{align} Now \begin{align}\cos \left( u+v \right)&=\cos u\cos v-\sin ... \tan \left( u-v \right)$ ====Solution==== Given \begin{align}\sin u&=\dfrac{3}{5},\,\,\,\, 0\le u\le
- Question 1, Exercise 10.1
- }}\sin {{22}^{\circ }}$ ==== Solution ==== As \begin{align} \sin (\alpha +\beta )=\sin \alpha \cos \be... +\cos \alpha \sin \beta, \end{align} Therefore \begin{align} \sin {{37}^{\circ }}\cos {{22}^{\circ }}+\... rc }}\sin {{53}^{\circ }}$ ====Solution==== As \begin{align}\cos (\alpha -\beta )&=\cos \alpha \cos \be... a +\sin \alpha \sin \beta,\end{align} Therefore \begin{align}\cos {{83}^{\circ }}\cos {{53}^{\circ }}+\s
- Question, Exercise 10.1
- 3rd quadrant and $\cos$ is -ive in 3rd quadrant, \begin{align}\cos\alpha &=-\sqrt{1-\sin^2\alpha}\\ &=-\... 2nd quadrant and $\sin$ is +ive in 2nd quadrant, \begin{align}&=\sqrt{1-{{\left(-\dfrac{12}{13} \right)}^... \quad \sin \beta &=\frac{5}{13}.\end{align} Now \begin{align}\sin (\alpha-\beta )&=\sin \alpha\cos \beta... 3rd quadrant and $\cos$ is -ive in 3rd quadrant, \begin{align}\cos\alpha &=-\sqrt{1-\sin^2\alpha}\\ &=-\
- Question 2, Exercise 10.2
- the second quadrant, value of $\cos$ is negative \begin{align}\cos\theta &=-\sqrt{1-{{\sin }^{2}}\theta }... the following by using double angle identities: \begin{align}\sin 2\theta &=2\sin \theta \cos \theta \\ ... the second quadrant, value of $\cos$ is negative \begin{align}\cos\theta &=-\sqrt{1-{{\sin }^{2}}\theta }... the following by using double angle identities: \begin{align}\cos 2\theta &={{\cos }^{2}}\theta -{{\sin
- Question 2, Exercise 10.1
- pi }{3}-\dfrac{\pi }{4}$ and using the identity: \begin{align}\sin (\alpha -\beta )=\sin \alpha \cos \beta -\cos \alpha \sin.\end{align} \begin{align} \Rightarrow \quad \sin \left( \frac{\pi }{... irc }}+{{30}^{\circ }}$ and using the identity: \begin{align}\tan (\alpha +\beta )&=\dfrac{\tan \alpha +... c }}+{{45}^{{}^\circ }}$ and using the identity: \begin{align}\tan ({{60}^{\circ }}+{{45}^{\circ }})&=\df
- Question11 and 12, Exercise 10.1
- $ and $\gamma$ are angles of triangle, therefore \begin{align}&\alpha+\beta +\gamma =180^\circ\\ \implie... ta}{2}=90^\circ-\dfrac{\gamma}{2}\end{align} Now \begin{align}&\tan\left( \dfrac{\alpha }{2}+\dfrac{\beta... amma}{2}\right)=\cot\tfrac{\gamma}{2}\end{align} \begin{align} \implies & \frac{\tan\dfrac{\alpha}{2}\tan... and $\gamma $ are angles of triangle, therefore \begin{align}&\alpha +\beta +\gamma =180^\circ\\ \impli
- Question 6, Exercise 10.2
- {{15}^{\circ }}$by using half angle identity as, \begin{align}\cos {{15}^{\circ }}&=\cos \dfrac{{{30}^{\c... 67.5}^{\circ }}$by using half angle identity as, \begin{align}\tan {{67.5}^{\circ }}&=\tan \dfrac{{{135}^... 12.5}^{\circ }}$by using half angle identity as, \begin{align} sin{{112.5}^{\circ }}&=\sin \dfrac{{{225}... \dfrac{\pi }{8}$by using half angle identity as, \begin{align} \cos \dfrac{\pi }{8}&=\cos \dfrac{\dfrac{
- Question 8, Exercise 10.1
- }{\cos \theta -\sin \theta }$ ====Solution==== \begin{align}L.H.S.&=\tan \left( \dfrac{\pi }{4}+\theta ... {1-tan\theta }{1+tan\theta }$ ====Solution==== \begin{align}L.H.S.&=\tan \left( \dfrac{\pi }{4}-\theta ... \ &=R.H.S.\end{align} ===Alternative Method=== \begin{align}L.H.S.&=\tan\left( \dfrac{\pi }{4}-\theta \... }\alpha {{\tan }^{2}}\beta }$ ====Solution==== \begin{align}\tan (\alpha +\beta )&=\dfrac{\tan \alpha +
- Question 4 and 5, Exercise 10.2
- of $\theta$ is in the third quadrant, that is, \begin{align}&\pi < \theta < \dfrac{3\pi}{2} \\ \implies... nd $\sin$ is positive in 2nd quadrant, therefore \begin{align}\sin\dfrac{\theta }{2}&=-\sqrt{\dfrac{1-\co... 3}$.\\ By using double angle identities, we have \begin{align}\sin 2\theta &=2\sin \theta \cos \theta \\ ... 3}$ \\ By using double angle identities, we have \begin{align}\cos 2\theta &=2{{\cos }^{2}}\theta -1\\ \
- Question 8 and 9, Exercise 10.2
- f one or more cosine functions. ====Solution==== \begin{align}{{\cos}^{4}}\theta &={{\left( {{\cos }^{2}}... a -4\sin \theta \cos \theta $ . ====Solution==== \begin{align}L.H.S.&=\sin 4\theta \\ &=\sin 2\left( 2\th... 2}}2\theta }{2\tan 2\theta }$ . ====Solution==== \begin{align}L.H.S.&=\cot 4\theta =\dfrac{\cos 4\theta }... cot\theta}{3\cot^2\theta -1}$ . ====Solution==== \begin{align}L.H.S.&=\cot 3\theta =\dfrac{1}{\tan3\theta
- Question 1, Exercise 10.3
- a -\beta ).$$ Put $\alpha =6x$ and $\beta =x$ \begin{align}-\,2\sin 6x\sin x&=\cos (6x+x)-\cos (6x-x)\... {{55}^{\circ }}$ and $\beta ={{123}^{\circ }}$ \begin{align}2\sin {{55}^{\circ }}\cos {{123}^{\circ }}&... a =\dfrac{A+B}{2}$ and $\beta =\dfrac{A-B}{2}$ \begin{align}& 2\sin \dfrac{A+B}{2}\cos \dfrac{A-B}{2}\\... pha =\dfrac{P+Q}{2}$ and $\beta =\dfrac{P-Q}{2}$ \begin{align}& 2\cos \dfrac{P+Q}{2}\cos \dfrac{P-Q}{2}\\
- Question 2, Exercise 10.3
- lpha ={{37}^{\circ }}$, $\beta ={{43}^{\circ }}$ \begin{align}\sin {{37}^{\circ }}+\sin {{43}^{\circ }}&=... $$ Put $\alpha =36^\circ$, adn $\beta =82^\circ$ \begin{align}\cos {{36}^{\circ }}-\cos {{82}^{\circ }}&=... pha =\dfrac{P+Q}{2}$ and $\beta =\dfrac{P-Q}{2}$ \begin{align}&\sin \dfrac{P+Q}{2}-\sin \dfrac{P-Q}{2}\\ ... pha =\dfrac{A+B}{2}$ and $\beta =\dfrac{A-B}{2}$ \begin{align}&\cos \dfrac{A+B}{2}+\cos \dfrac{A-B}{2}\\