Question

Prave that:

(1) $\frac{{\sin }^2\mathrm{\theta }}{\mathrm{cos\theta }}+\mathrm{cos\theta }=\mathrm{sec\theta }$

(2) ${\cos }^2\mathrm{\theta }\left(1+{\tan }^2\mathrm{\theta }\right)=1$

(3) $\sqrt{\frac{1-\mathrm{sin\theta }}{1+\mathrm{sin\theta }}}=\mathrm{sec\theta }-\mathrm{tan\theta }$

(4) $\left(\mathrm{sec\theta }-\mathrm{cos\theta }\right)\left(\mathrm{cot\theta }+\mathrm{tan\theta }\right)=\mathrm{tan\theta }\mathrm{sec\theta }$

(5) $\mathrm{cot\theta }+\mathrm{tan\theta }=\mathrm{cosec\theta }\mathrm{sec\theta }$

(6) $\frac{1}{\mathrm{sec\theta }-\mathrm{tan\theta }}=\mathrm{sec\theta }+\mathrm{tan\theta }$

(7) ${\sec }^4\mathrm{\theta }-{\cos }^4\mathrm{\theta }=1-2{\cos }^2\mathrm{\theta }$

(8) $\mathrm{sec\theta }+\mathrm{tan\theta }=\frac{\cos {\displaystyle \mathrm{\theta }}}{1-\mathrm{sin\theta }}$

(9) If $t\mathrm{an\theta }+\frac{1}{\mathrm{tan\theta }}=2$, then show that ${\tan }^2\mathrm{\theta }+\frac{1}{{\tan }^2\mathrm{\theta }}=2$

(10) $\frac{\mathrm{tanA}}{{\left(1+{\tan }^2\mathrm{A}\right)}^2}+\frac{\mathrm{cotA}}{{\left(1+{\cot }^2\mathrm{A}\right)}^2}=\sin \mathrm{A}\cos \mathrm{A}$

(11) ${\sec }^4\mathrm{A}\left(1-{\sin }^4\mathrm{A}\right)-2{\tan }^2\mathrm{A}=1$

(12) $\frac{\mathrm{tan\theta }}{\mathrm{sec\theta }-1}=\frac{\mathrm{tan\theta }+\mathrm{sec\theta }+1}{\mathrm{tan\theta }+\mathrm{sec\theta }-1}$

Answer 1

(1)
$$\begin{gathered} \frac{{\sin }^2\theta }{\cos \theta }+\cos \theta \\ =\frac{{\displaystyle {\sin }^2}{\displaystyle \theta }{\displaystyle +}{\displaystyle {{\displaystyle \cos }}^2}{\displaystyle \theta }}{\cos {\displaystyle \theta }} \\ =\frac{1}{\cos {\displaystyle \theta }} \\ =\sec \theta \end{gathered}$$
(2)
$$\begin{gathered} {\cos }^2\theta \left(1+{\tan }^2\theta \right) \\ ={\cos }^2\theta \times {\sec }^2\theta \\ ={\cos }^2\theta \times \frac{1}{{\cos }^2\theta } \\ =1 \end{gathered}$$
(3)
$$\begin{gathered} \sqrt{\frac{1-\sin \theta }{1+\sin \theta }} \\ =\sqrt{\frac{1-\sin \theta }{1+\sin \theta }\times \frac{1-\sin \theta }{1-\sin \theta }} \\ =\sqrt{\frac{{\left(1-\sin \theta \right)}^2}{1-{\sin }^2\theta }} \\ =\sqrt{\frac{{\left(1-\sin \theta \right)}^2}{{\cos }^2\theta }}\left({\cos }^2\theta +{\sin }^2\theta =1\right) \end{gathered}$$
$$\begin{gathered} =\frac{1-\sin \theta }{\cos \theta } \\ =\frac{1}{\cos \theta }-\frac{\sin \theta }{\cos \theta } \\ =\sec \theta -\tan \theta \end{gathered}$$
(4)
$$\begin{gathered} \left(\sec \theta -\cos \theta \right)\left(\cot \theta +\tan \theta \right) \\ =\left(\frac{1}{\cos \theta }-\cos \theta \right)\left(\frac{\cos \theta }{\sin \theta }+\frac{\sin \theta }{\cos \theta }\right) \\ =\left(\frac{1-{\cos }^2\theta }{\cos \theta }\right)\left(\frac{{\sin }^2\theta +{\cos }^2\theta }{\sin \theta \cos \theta }\right) \\ =\frac{{\sin }^2\theta }{\cos \theta }\times \frac{1}{\sin \theta \cos \theta }\left({\sin }^2\theta +{\cos }^2\theta =1\right) \\ =\frac{\sin \theta }{\cos \theta }\times \frac{1}{\cos \theta } \\ =\tan \theta \sec \theta \end{gathered}$$
(5)
$$\begin{gathered} \cot \theta +\tan \theta \\ =\frac{\cos \theta }{\sin \theta }+\frac{\sin \theta }{\cos \theta } \\ =\frac{{\sin }^2\theta +{\cos }^2\theta }{\sin \theta \cos \theta } \\ =\frac{1}{\sin \theta \cos \theta }\left({\sin }^2\theta +{\cos }^2\theta =1\right) \\ =\frac{1}{\sin \theta }\times \frac{1}{\cos \theta } \\ =\mathrm{cosec}\theta \sec \theta \end{gathered}$$
(6)
$$\begin{gathered} \frac{1}{\sec \theta -\tan \theta } \\ =\frac{{\displaystyle 1}}{{\displaystyle \sec \theta -\tan \theta }}\times \frac{\sec \theta +\tan \theta }{\sec \theta +\tan \theta } \\ =\frac{\sec \theta +\tan \theta }{{\sec }^2\theta -{\tan }^2\theta }\left[\left(a+b\right)\left(a-b\right)=a^2-b^2\right] \\ =\sec \theta +\tan \theta \left(1+{\tan }^2\theta ={\sec }^2\theta \right) \end{gathered}$$
(7)
Disclaimer: There is printing mistake in the question. The correct question should be ${\sin }^4\mathrm{\theta }-{\cos }^4\mathrm{\theta }=1-2{\cos }^2\mathrm{\theta }$. The solution has been provided accordingly.

$$\begin{gathered} {\sin }^4\theta -{\cos }^4\theta \\ ={\left({\sin }^2\theta \right)}^2-{\left({\cos }^2\theta \right)}^2 \\ =\left({\sin }^2\theta -{\cos }^2\theta \right)\left({\sin }^2\theta +{\cos }^2\theta \right)\left[a^2-b^2=\left(a+b\right)\left(a-b\right)\right] \\ ={\sin }^2\theta -{\cos }^2\theta \left({\sin }^2\theta +{\cos }^2\theta =1\right) \\ =1-{\cos }^2\theta -{\cos }^2\theta \\ =1-2{\cos }^2\theta \end{gathered}$$
(8)
$$\begin{gathered} \sec \theta +\tan \theta \\ =\frac{1}{\cos \theta }+\frac{\sin \theta }{\cos \theta } \\ =\frac{1+\sin \theta }{\cos \theta } \\ =\frac{\cos \theta \left(1+\sin \theta \right)}{{\cos }^2\theta } \\ =\frac{\cos \theta \left(1+\sin \theta \right)}{1-{\sin }^2\theta }\left({\sin }^2\theta +{\cos }^2\theta =1\right) \end{gathered}$$
$$\begin{gathered} =\frac{\cos {\displaystyle \theta }{\displaystyle \left(1+\sin \theta \right)}}{\left(1+\sin \theta \right){\displaystyle \left(1-\sin \theta \right)}} \\ =\frac{{\displaystyle \cos \theta }}{{\displaystyle 1-\sin \theta }} \end{gathered}$$
(9)
$\tan \theta +\frac{1}{\tan \theta }=2$

Squaring on both sides, we get

$$\begin{gathered} {\left(\tan \theta +\frac{1}{\tan \theta }\right)}^2=2^2 \\ \Rightarrow {\tan }^2\theta +\frac{1}{{\tan }^2\theta }+2\times \tan \theta \times \frac{1}{\tan \theta }=4 \\ \Rightarrow {\tan }^2\theta +\frac{1}{{\tan }^2\theta }+2=4 \\ \Rightarrow {\tan }^2\theta +\frac{1}{{\tan }^2\theta }=4-2=2 \end{gathered}$$
(10)
$$\begin{gathered} \frac{\tan A}{{\left(1+{\tan }^{2}A\right)}^2}+\frac{\cot A}{{\displaystyle {\left(1+{\cot }^{2}A\right)}^2}} \\ =\frac{{\displaystyle \tan A}}{{\displaystyle {\left({\sec }^{2}A\right)}^2}}+\frac{{\displaystyle \cot A}}{{\displaystyle {\left({\mathrm{cosec}}^{2}A\right)}^2}}\left(1+{\tan }^2\theta ={\sec }^2\theta \mathrm{and}1+{\cot }^2\theta ={\mathrm{cosec}}^2\theta \right) \\ =\frac{\sin {\displaystyle A}}{\cos {\displaystyle A}}\times {\cos }^{4}A+\frac{\cos {\displaystyle A}}{\sin {\displaystyle A}}\times {\sin }^{4}A\left(\cos \theta =\frac{1}{\sec \theta }\mathrm{and}\sin \theta =\frac{1}{\cos {\displaystyle \mathrm{ec}}{\displaystyle \theta }}\right) \\ =\sin A{\cos }^{3}A+\cos A{\sin }^{3}A \end{gathered}$$
$$\begin{gathered} =\sin A\cos A\left({\cos }^{2}A+{\sin }^{2}A\right) \\ =\sin A\cos A\left({\cos }^2\theta +{\sin }^2\theta =1\right) \end{gathered}$$
(11)
$$\begin{gathered} {\sec }^{4}A\left(1-{\sin }^{4}A\right)-2{\tan }^{2}A \\ ={\sec }^{4}A-{\sec }^{4}A{\sin }^{4}A-2{\tan }^{2}A \\ ={\left(1+{\tan }^{2}A\right)}^2-\frac{{\sin }^{4}A}{{\cos }^{4}A}-2{\tan }^{2}A\left({\sec }^2\theta =1+{\tan }^2\theta \mathrm{and}\sec \theta =\frac{1}{\cos \theta }\right) \\ =1+{\tan }^{4}A+2{\tan }^{2}A-{\tan }^{4}A-2{\tan }^{2}A\left[{\left(a+b\right)}^2=a^2+b^2+2ab\right] \\ =1 \end{gathered}$$
(12)
$$\begin{gathered} \frac{\tan \theta +\sec \theta +1}{\tan \theta +\sec \theta -1} \\ =\frac{{\displaystyle \tan \theta +\sec \theta +1}}{{\displaystyle \tan \theta +\sec \theta -1}}\times \frac{{\displaystyle \tan \theta +\sec \theta +1}}{{\displaystyle \tan \theta +\sec \theta +1}} \\ =\frac{{\displaystyle {\tan }^2\theta +{\sec }^2\theta +1+2\tan \theta +2\sec \theta +2\sec \theta \tan \theta }}{{\displaystyle {\tan }^2\theta +{\sec }^2\theta +2\tan \theta \sec \theta -1}} \\ =\frac{{\displaystyle 2{\sec }^2\theta +2\tan \theta +2\sec \theta +2\sec \theta \tan \theta }}{{\displaystyle 2{\tan }^2\theta +2\tan \theta \sec \theta }}\left(1+{\tan }^2\theta ={\sec }^2\theta \right) \\ =\frac{{\displaystyle 2\left(\tan \theta +\sec \theta \right)+2\sec \theta \left(\tan \theta +\sec \theta \right)}}{{\displaystyle 2\tan \theta \left(\tan \theta +\sec \theta \right)}} \end{gathered}$$
$$\begin{gathered} =\frac{2\left(\tan \theta +\sec \theta \right)\left(\sec \theta +1\right)}{2\tan \theta \left(\tan \theta +\sec \theta \right)} \\ =\frac{\sec \theta +1}{\tan \theta } \\ =\frac{\sec \theta +1}{\tan \theta }\times \frac{\sec \theta -1}{\sec \theta -1} \\ =\frac{{\sec }^2\theta -1}{\tan \theta \left(\sec \theta -1\right)}\left[\left(a-b\right)\left(a+b\right)=a^2-b^2\right] \\ =\frac{{\tan }^2\theta }{\tan \theta \left(\sec \theta -1\right)} \\ =\frac{\tan \theta }{\sec \theta -1} \end{gathered}$$