Question

Prove that
(i) sin (70° + θ) − cos (20° − θ) = 0
(ii) tan (55° − θ) − cot (35° + θ) = 0
(iii) cosec (67° + θ) − sec (23° − θ) = 0
(iv) cosec (65 °+ θ) sec (25° − θ) − tan (55° − θ) + cot (35° + θ) = 0
(v) sin (50° + θ ) − cos (40° − θ) + tan 1° tan 10° tan 80° tan 89° = 1.

Answer 1

$$\begin{gathered} \begin{array}{l}\\ \left(\text{i}\right)\text{L}\text{.H}\text{.S=sin}({70}^0+\theta )-\cos ({20}^0-\theta )\\ \end{array} \\ \begin{array}{l}\text{=sin}\{{90}^0-({20}^0-\theta )\}-\cos ({20}^0-\theta )\\ \end{array} \\ \begin{array}{l}\text{=}\cos ({20}^0-\theta )-\cos ({20}^0-\theta )\\ \end{array} \\ \begin{array}{l}\text{=0}\\ \begin{array}{l}\text{=R}\text{.H}\text{.S}\text{.}\\ \end{array}\end{array} \\ \begin{array}{l}\left(\text{ii}\right)\text{L}\text{.H}\text{.S=tan}({55}^0-\theta )-\cot ({35}^0+\theta )\\ \end{array} \\ \begin{array}{l}\text{=tan}\{{90}^0-({35}^0+\theta )\}-\cot ({35}^0+\theta )\\ \end{array} \\ \begin{array}{l}\text{=}\cot ({35}^0+\theta )-\cot ({35}^0+\theta )\\ \end{array} \\ \begin{array}{l}\text{=0=R}\text{.H}\text{.S}\text{.}\\ \end{array} \\ \begin{array}{l}\left(\text{iii}\right)\text{L}\text{.H}\text{.S=cosec}({67}^0+\theta )-\sec ({23}^0-\theta )\\ \end{array} \\ \begin{array}{l}\text{=cosec}\{{90}^0-({23}^0-\theta )\}-\sec ({23}^0-\theta )\\ \end{array} \\ \begin{array}{l}\text{=}\sec ({23}^0-\theta )-\sec ({23}^0-\theta )\\ \\ \end{array} \\ \text{=0=R}\text{.H}\text{.S}\text{.} \\ \begin{array}{l}\\ \left(\text{iv}\right)\text{L}\text{.H}\text{.S=cosec}({65}^0+\theta )-\sec ({25}^0-\theta )-\tan ({55}^0-\theta )+\cot ({35}^0+\theta )\\ \end{array} \\ \begin{array}{l}=\cos ec\{{90}^0-({25}^0-\theta )\}-\sec ({25}^0-\theta )-\tan ({55}^0-\theta )+\cot \{{90}^0-({55}^0-\theta )\}\\ \end{array} \\ \begin{array}{l}=\sec ({25}^0-\theta )-\sec ({25}^0-\theta )-\tan ({55}^0-\theta )+\tan ({55}^0-\theta )\\ \end{array} \\ \begin{array}{l}=0=\text{R}\text{.H}\text{.S}\\ \end{array} \\ \begin{array}{l}\left(\text{v}\right)\text{L}\text{.H}\text{.S}=\sin ({50}^0+\theta )-\cos ({40}^0-\theta )+\tan 1^0\tan {10}^0\tan {80}^0\tan {89}^0\\ \end{array} \\ \begin{array}{l}=\sin \{{90}^0-({40}^0-\theta )\}-\cos ({40}^0-\theta )+\left\{\tan 1^0\tan ({90}^0-1^0)\right\}\left\{\tan {10}^0\tan ({90}^0-{10}^{})\right\}\\ \end{array} \\ \begin{array}{l}=\cos ({40}^0-\theta )-\cos ({40}^0-\theta )+\left(\tan 1^0\cot 1^0\right)\left(\tan {10}^0\cot {10}^0\right)\\ \end{array} \\ \begin{array}{l}=(\frac{1}{\cot 1^0}\times \cot 1^0)(\tan {10}^0\times \frac{1}{\tan {10}^0})\\ \end{array} \\ \begin{array}{l}=1\times 1\\ \end{array} \\ =1=\text{R}\text{.H}\text{.S} \end{gathered}$$