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Prove that: (i) sin (70^@+theta) -cos (20^@-theta) =0 (ii) tan (55^@-theta) -cot (35^@-theta)=0 (iii) cosec(67^@+theta) -sec (23^@-theta)=0 (iv) cosec (65^@+theta) -sec (25^@-theta )-tan (55^@-theta) +cot (35^@+theta) =0 (v) sin (50^@+theta) -cos (40^@-theta) +tan 1^@ tan 10 ^@ tan 80^@ tan 89^@=1 |
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Answer» Solution :`(i) LHS =SIN {90^@-(20^@-theta)}-cos (20^@-theta)=cos (20^@-theta) -cos (20^@-theta) =0`. `(v) LHS =sin {90^@-(40^@-theta )}-cos (40^@-theta)+(tan 1^@tan 89^@)(tan 10^@tan 80^@)=cos (40^@-theta) -cos (40^@-theta) +(1xx1)=1`. |
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