Question

Prove that:

(i) $\frac{sin{70}^{\circ }}{cos{20}^{\circ }}+\frac{cosec{20}^{\circ }}{sec{70}^{\circ }}-2cos{70}^{\circ }cosec{20}^{\circ }=0$

(ii) $\frac{cos{80}^{\circ }}{sin{10}^{\circ }}+cos{59}^{\circ }cosec{31}^{\circ }=2$

(iii) $\frac{2sin{68}^{\circ }}{cos{22}^{\circ }}-\frac{2cot{15}^{\circ }}{5tan{75}^{\circ }}-\frac{3tan{45}^{\circ }tan{20}^{\circ }tan{40}^{\circ }tan{50}^{\circ }tan{70}^{\circ }}{5}=1$

(iv) $\frac{sin{18}^{\circ }}{cos{72}^{\circ }}+\sqrt{3}\left(tan{10}^{\circ }tan{30}^{\circ }tan{40}^{\circ }tan{50}^{\circ }tan{80}^{\circ }\right)=2$

(v) $\frac{7cos{55}^{\circ }}{3sin{35}^{\circ }}-\frac{4\left(cos{70}^{\circ }cosec{20}^{\circ }\right)}{3\left(tan{5}^{\circ }tan{25}^{\circ }tan{45}^{\circ }tan{65}^{\circ }tan{85}^{\circ }\right)}=1$

Answer 1

(i) $\frac{sin{70}^{\circ }}{cos{20}^{\circ }}+\frac{cosec{20}^{\circ }}{sec{70}^{\circ }}-2cos{70}^{\circ }cosec{20}^{\circ }=0LHS=\frac{sin{70}^o}{cos{20}^o}+\frac{cosec{20}^o}{sec{70}^o}–2cos{70}^{o}cosec{20}^o=\frac{sin{70}^o}{sin({90}^o–{20}^o)}+\frac{sec({90}^o-{20}^o)}{sec{70}^o}–2cos{70}^{o}sec({90}^o–{20}^o)=\frac{sin{70}^o}{sin{70}^o}+\frac{sec{70}^o}{sec{70}^o}-2cos{70}^{o}sec{70}^o=1+1-2\times cos{70}^o\times \frac{1}{cos{70}^o}=2-2=0=RHS$

(ii) $\frac{cos{80}^{\circ }}{sin{10}^{\circ }}+cos{59}^{\circ }cosec{31}^{\circ }=2LHS=\frac{cos{80}^o}{sin{10}^o}+cos{59}^{o}cosec{31}^o=\frac{cos{80}^o}{cos({90}^o–{10}^o)}+sin({90}^o–{59}^o)cosec{31}^o=\frac{cos{80}^o}{cos{80}^o}+sin{31}^{o}cosec{31}^o=1+sin{31}^o\times \frac{1}{sin{31}^o}=1+1=2=RHS$

(iii) $\frac{2sin{68}^{\circ }}{cos{22}^{\circ }}-\frac{2cot{15}^{\circ }}{5tan{75}^{\circ }}-\frac{3tan{45}^{\circ }tan{20}^{\circ }tan{40}^{\circ }tan{50}^{\circ }tan{70}^{\circ }}{5}=1LHS=\frac{2sin{68}^{\circ }}{cos{22}^{\circ }}-\frac{2cot{15}^{\circ }}{5tan{75}^{\circ }}-\frac{3tan{45}^{\circ }tan{20}^{\circ }tan{40}^{\circ }tan{50}^{\circ }tan{70}^{\circ }}{5}=\frac{2sin{68}^o}{sin({90}^o–{22}^o)}-\frac{2cot{15}^o}{5cot({90}^o–{75}^o)}–\frac{3\times 1\times cot({90}^o–{20}^o)\times cot({90}^o–{40}^o)\times tan{50}^o\times tan{70}^o}{5}=\frac{2sin{68}^o}{sin{68}^o}–\frac{2cot{15}^o}{5cot{15}^o}–\frac{3cot{70}^{o}cot{50}^{o}tan{50}^{o}tan{70}^o}{5}=2–\frac{2}{5}–\frac{3\times \frac{1}{tan{70}^o}\times \frac{1}{tan{50}^o}\times tan{50}^o\times tan{70}^o}{5}=2-\frac{2}{5}-\frac{3}{5}=\frac{10–2–3}{5}=\frac{5}{5}=1=RHS$

(iv) $\frac{sin{18}^{\circ }}{cos{72}^{\circ }}+\sqrt{3}\left(tan{10}^{\circ }tan{30}^{\circ }tan{40}^{\circ }tan{50}^{\circ }tan{80}^{\circ }\right)=2LHS=\frac{sin{18}^{\circ }}{cos{72}^{\circ }}+\sqrt{3}\left(tan{10}^{\circ }tan{30}^{\circ }tan{40}^{\circ }tan{50}^{\circ }tan{80}^{\circ }\right)=\frac{sin{18}^o}{sin({90}^o–{72}^o)}+\sqrt{3}\left[cot\right({90}^o-{10}^o)\times \frac{1}{\sqrt{3}}\times cot({90}^o–{40}^o)\times tan{50}^o\times tan{80}^o]=\frac{sin{18}^o}{sin{18}^o}+\frac{\sqrt{3}(cot{80}^o\times cot{50}^o\times tan{50}^o\times tan{80}^o)}{\sqrt{3}}=1+(\frac{1}{tan{80}^o}\times \frac{1}{tan{50}^o}\times tan{50}^o\times tan{80}^o)=1+1=2=RHS$

(v) $\frac{7cos{55}^{\circ }}{3sin{35}^{\circ }}-\frac{4\left(cos{70}^{\circ }cosec{20}^{\circ }\right)}{3\left(tan{5}^{\circ }tan{25}^{\circ }tan{45}^{\circ }tan{65}^{\circ }tan{85}^{\circ }\right)}=1LHS=\frac{7cos{55}^{\circ }}{3sin{35}^{\circ }}-\frac{4\left(cos{70}^{\circ }cosec{20}^{\circ }\right)}{3\left(tan{5}^{\circ }tan{25}^{\circ }tan{45}^{\circ }tan{65}^{\circ }tan{85}^{\circ }\right)}=\frac{7cos{55}^o}{3cos({90}^o–{35}^o)}–\frac{4\left(sin\right({90}^o–{70}^o\left)cosec{20}^o\right)}{3\left(cot\right({90}^o–5^o)\times cot({90}^o–{25}^o)\times 1\times tan{65}^o\times tan{85}^o)}=\frac{7cos{55}^o}{3cos{55}^o}–\frac{4\left(sin{20}^{o}cosec{20}^o\right)}{3\left(cot{85}^{o}cot{65}^{o}tan{65}^{o}tan{85}^o\right)}=\frac{7}{3}–\frac{4(sin{20}^o\times \frac{1}{sin{20}^o})}{3(\frac{1}{tan{85}^o}\times \frac{1}{tan{65}^o}\times tan{65}^o\times tan{85}^o)}=\frac{7}{3}–\frac{4}{3}=\frac{3}{3}=1=RHS$