SIMPLY :
Sin48 Sec42 + Cos48 Cosec42

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Solution

sin48.sec42+cos48.cosec42 =sin48/cos42+cos48.sin42 =sin48/cos(90-48)+cos48.sin(90-48) =sin48/sin48+cos48/cos48 =1+1 =2

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Sin48.sec42+cos48.cosec42= ........

sin 48^{\circ} sec 42^{\circ} + cos 48^{\circ} cosec 42^{\circ} = 2

Find the value of \sin 48 ° \sec 42 ° + \cos 48 ° cosec 42 ° .

Prove that:

(i) tan $20^{o} t a n 35^{o} t a n 45^{o} t a n 55^{o} t a n 70^{o} = 1$

(ii) sin $48^{o} s e c 42^{o} + c o s 48^{o} c o s e c 42^{o} = 2$

(iii) $\frac{s i n 70^{o}}{c o s 20^{o}} + \frac{c o s e c 20^{o}}{s e c 70^{o}} - 2 c o s 70^{o} c o s e c 20^{o} = 0$

(iv) $\frac{c o s 80^{o}}{s i n 10^{o}} + c o s 59^{o} c o s e c 31^{o} = 2$

Evaluate :
$cos 48^{\circ} - sin 42^{\circ}$ ___

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