Question

Prove that:

(i) $sin({70}^{\circ }+\theta )-cos({20}^{\circ }-\theta )=0$

(ii) $tan({55}^{\circ }-\theta )-cot({35}^{\circ }+\theta )=0$

(iii) $cosec({67}^{\circ }+\theta )-sec({23}^{\circ }-\theta )=0$

(iv) $cosec({65}^{\circ }+\theta )-sec({25}^{\circ }-\theta )-tan({55}^{\circ }-\theta )+cot({35}^{\circ }+\theta )=0$

(v) $sin({50}^{\circ }+\theta )-cos({40}^{\circ }-\theta )+tan{1}^{\circ }tan{10}^{\circ }tan{80}^{\circ }tan{89}^{\circ }=1$

Answer 1

(i) $sin({70}^o+\theta )-cos({20}^0-\theta )=0LHS=sin({70}^o+\theta )-cos({20}^0-\theta )sin({70}^o+\theta )-sin({90}^o-({20}^0-\theta \left)\right)sin({70}^o+\theta )-sin({70}^o+\theta )=0=RHS$

(ii) $tan({55}^o-\theta )-cot({35}^o+\theta )=0LHS=tan({55}^o-\theta )-cot({35}^o+\theta )=cot({90}^o-({55}^o-\theta \left)\right)-cot({35}^o+\theta )=cot({35}^o+\theta )-cot({35}^o+\theta )=0=RHS$

(iii)$cosec({67}^{\circ }+\theta )-sec({23}^{\circ }-\theta )=0=0LHS=cosec({67}^{\circ }+\theta )-sec({23}^{\circ }-\theta )=0=sec({90}^o-({67}^o+\theta \left)\right)-sec({23}^o-\theta )=sec({23}^o-\theta )-sec({23}^o-\theta )=0=RHS$

(iv) $cosec({65}^{\circ }+\theta )-sec({25}^{\circ }-\theta )-tan({55}^{\circ }-\theta )+cot({35}^{\circ }+\theta )=0LHS=cosec({65}^{\circ }+\theta )-sec({25}^{\circ }-\theta )-tan({55}^{\circ }-\theta )+cot({35}^{\circ }+\theta )=sec({90}^o-({65}^o+\theta \left)\right)-sec({25}^o-\theta )-cot({90}^o-({55}^o-\theta \left)\right)-cot({35}^o+\theta )=sec({25}^o-\theta )-sec({25}^o-\theta )-cot({35}^o+\theta )+cot({35}^o+\theta )=0RHS$

(v) $sin({50}^{\circ }+\theta )-cos({40}^{\circ }-\theta )+tan{1}^{\circ }tan{10}^{\circ }tan{80}^{\circ }tan{89}^{\circ }=1LHS=sin({50}^{\circ }+\theta )-cos({40}^{\circ }-\theta )+tan{1}^{\circ }tan{10}^{\circ }tan{80}^{\circ }tan{89}^{\circ }=cos({90}^o-({50}^{\circ }+\theta \left)\right)-cos({40}^{\circ }-\theta )+tan{1}^{\circ }tan{10}^{\circ }tan({90}^o-{80}^o)tan({90}^o-{89}^o)=cos({40}^{\circ }-\theta )-cos({40}^{\circ }-\theta )+tan{1}^{\circ }tan{10}^{\circ }cot{1}^{\circ }cot{10}^{\circ }=0+tan{1}^{\circ }tan{10}^{\circ }\times \frac{1}{tan{1}^{\circ }}\times \frac{1}{tan{10}^{\circ }}=0+1=1=RHS$