Question

Prove the following.

(1) secθ (1 – sinθ) (secθ + tanθ) = 1

(2) (secθ + tanθ) (1 – sinθ) = cosθ

(3) sec²θ + cosec²θ = sec²θ × cosec²θ

(4) cot²θ – tan²θ = cosec²θ – sec²θ

(5) tan⁴θ + tan²θ = sec⁴θ – sec²θ

(6) $\frac{1}{1-\sin \theta }+\frac{{\displaystyle 1}}{{\displaystyle 1+\sin \theta }}=2{\sec }^2\theta$

(7) sec⁶x – tan⁶x = 1 + 3sec²x × tan²x

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

(9) $\frac{{\tan }^3\theta -1}{\tan \theta -1}={\sec }^2\theta +\tan \theta$

(10) $\frac{\sin \theta -\cos \theta +1}{\sin \theta +\cos \theta -1}=\frac{1}{\sin \theta -\tan \theta }$

Answer 1

(1)
$$\begin{gathered} \sec \theta \left(1-\sin \theta \right)\left(\sec \theta +\tan \theta \right) \\ =\left(\sec \theta -\sec \theta \sin \theta \right)\left(\sec \theta +\tan \theta \right) \\ =\left(\sec \theta -\frac{\sin \theta }{\cos \theta }\right)\left(\sec \theta +\tan \theta \right) \\ =\left(\sec \theta -\tan \theta \right)\left(\sec \theta +\tan \theta \right) \\ ={\sec }^2\theta -{\tan }^2\theta \\ =1\left(1+{\tan }^2\theta ={\sec }^2\theta \right) \end{gathered}$$
(2)
$$\begin{gathered} \left(\sec \theta +\tan \theta \right)\left(1-\sin \theta \right) \\ =\left(\frac{1}{\cos \theta }+\frac{\sin \theta }{\cos \theta }\right)\left(1-\sin \theta \right) \\ =\left(\frac{1+\sin \theta }{\cos \theta }\right)\left(1-\sin \theta \right) \\ =\frac{1-{\sin }^2\theta }{\cos \theta } \\ =\frac{{\cos }^2\theta }{\cos \theta }\left({\sin }^2\theta +{\cos }^2\theta =1\right) \\ =\cos \theta \end{gathered}$$
(3)
$$\begin{gathered} {\sec }^2\theta +{\mathrm{cosec}}^2\theta \\ =\frac{1}{{\cos }^2\theta }+\frac{1}{{\sin }^2\theta } \\ =\frac{{\sin }^2\theta +{\cos }^2\theta }{{\cos }^2\theta {\sin }^2\theta } \\ =\frac{1}{{\cos }^2\theta {\sin }^2\theta } \\ =\frac{1}{{\cos }^2\theta }\times \frac{1}{{\sin }^2\theta } \\ ={\sec }^2\theta {\mathrm{cosec}}^2\theta \end{gathered}$$
(4)
$$\begin{gathered} {\cot }^2\theta -{\tan }^2\theta \\ =\left({\mathrm{cosec}}^2\theta -1\right)-\left({\sec }^2\theta -1\right)\left(1+{\tan }^2\theta ={\sec }^2\theta \&1+{\cot }^2\theta ={\mathrm{cosec}}^2\theta \right) \\ ={\mathrm{cosec}}^2\theta -1-{\sec }^2\theta +1 \\ ={\mathrm{cosec}}^2\theta -{\sec }^2\theta \end{gathered}$$
(5)
$$\begin{gathered} {\tan }^4\theta +{\tan }^2\theta \\ ={\left({\sec }^2\theta -1\right)}^2+\left({\sec }^2\theta -1\right)\left(1+{\tan }^2\theta ={\sec }^2\theta \right) \\ ={\sec }^4\theta -2{\sec }^2\theta +1+{\sec }^2\theta -1\left[{\left(a-b\right)}^2=a^2-2ab+b^2\right] \\ ={\sec }^4\theta -{\sec }^2\theta \end{gathered}$$
(6)
$$\begin{gathered} \frac{1}{1-\sin \theta }+\frac{1}{1+\sin \theta } \\ =\frac{{\displaystyle 1+\sin \theta +1-\sin \theta }}{{\displaystyle \left(1-\sin \theta \right)\left(1+\sin \theta \right)}} \\ =\frac{2}{1-{\sin }^2\theta }\left[\left(a-b\right)\left(a+b\right)=a^2-b^2\right] \\ =\frac{{\displaystyle 2}}{{\displaystyle {\cos }^2\theta }}\left({\sin }^2\theta +{\cos }^2\theta =1\right) \\ =2{\sec }^2\theta \end{gathered}$$
(7)
We have,

${\sec }^{2}x-{\tan }^{2}x=1$

Cubing on both sides, we get

$$\begin{gathered} {\left({\sec }^{2}x-{\tan }^{2}x\right)}^3=1^3 \\ \Rightarrow {\left({\sec }^{2}x\right)}^3-{\left({\tan }^{2}x\right)}^3-3\times {\sec }^{2}x\times {\tan }^{2}x\times \left({\sec }^{2}x-{\tan }^{2}x\right)=1\left[{\left(a-b\right)}^3=a^3-b^3-3ab\left(a-b\right)\right] \\ \Rightarrow {\sec }^{6}x-{\tan }^{6}x-3{\sec }^{2}x{\tan }^{2}x=1 \\ \Rightarrow {\sec }^{6}x-{\tan }^{6}x=1+3{\sec }^{2}x{\tan }^{2}x \end{gathered}$$
(8)
$$\begin{gathered} \frac{\tan \theta }{\sec \theta +1} \\ =\frac{{\displaystyle \tan \theta }}{{\displaystyle \sec \theta +1}}\times \frac{\sec \theta -1}{\sec \theta -1} \\ =\frac{\tan {\displaystyle \theta }{\displaystyle \left(\sec \theta -1\right)}}{{\sec }^2\theta -1} \\ =\frac{{\displaystyle \tan \theta \left(\sec \theta -1\right)}}{{\tan }^2\theta }\left(1+{\tan }^2\theta ={\sec }^2\theta \right) \\ =\frac{{\displaystyle \sec \theta -1}}{\tan \theta } \end{gathered}$$
(9)
$$\begin{gathered} \frac{{\tan }^3\theta -1}{\tan \theta -1} \\ =\frac{{\displaystyle \left(\tan \theta -1\right)\left({\tan }^2\theta +\tan \theta \times 1+1\right)}}{{\displaystyle \tan \theta -1}}\left[a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)\right] \\ ={\tan }^2\theta +\tan \theta +1 \\ ={\sec }^2\theta +\tan \theta \left(1+{\tan }^2\theta ={\sec }^2\theta \right) \end{gathered}$$
(10)
$\frac{\sin \theta -\cos \theta +1}{\sin \theta +\cos \theta -1}$
$=\frac{{\displaystyle \frac{\sin \theta -\cos \theta +1}{\cos \theta }}}{{\displaystyle \frac{\sin \theta +\cos \theta -1}{\cos \theta }}}$ (Dividing numerator and denominator by cosθ)
$=\frac{{\displaystyle \frac{\sin \theta }{\cos \theta }}-{\displaystyle \frac{\cos \theta }{\cos \theta }}+{\displaystyle \frac{1}{\cos \theta }}}{{\displaystyle \frac{\sin \theta }{\cos \theta }}+{\displaystyle \frac{\cos \theta }{\cos \theta }}-{\displaystyle \frac{1}{\cos \theta }}}$
$$\begin{gathered} =\frac{\tan \theta -1+\sec \theta }{\tan \theta +1-\sec \theta } \\ =\frac{\tan \theta -1+\sec \theta }{\tan \theta +\left({\sec }^2\theta -{\tan }^2\theta \right)-\sec \theta }\left(1+{\tan }^2\theta ={\sec }^2\theta \right) \\ =\frac{\tan \theta -1+\sec \theta }{\left(\sec \theta -\tan \theta \right)\left(\sec \theta +\tan \theta \right)-\left(\sec \theta -\tan \theta \right)}\left[a^2-b^2=\left(a-b\right)\left(a+b\right)\right] \\ =\frac{\tan \theta -1+\sec \theta }{\left(\sec \theta -\tan \theta \right)\left(\sec \theta +\tan \theta -1\right)} \\ =\frac{1}{\sec \theta -\tan \theta } \end{gathered}$$