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Evaluate : (i) (sin18^@)/(cos72^@) ...

Evaluate :
(i) `(sin18^@)/(cos72^@)`
(ii) `(tan26^@)/(cot64^@)`
(iii) `cos48^@-sin42^@`
(iv) `c o s e c 31^@ - sec59^@`

Text Solution

Verified by Experts

(i) `sin 18^@/cos 72^@` Since, `sin (90^@ - θ) = cos θ`
Here, `θ = 72^@`
`sin 18^@/cos 72^@ = sin (90^@ - 72^@)/cos 72^@`
...
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