Question 89

-(1 + sec $$20^\circ$$ + cot $$70^\circ$$)(1 - cosec $$20^\circ$$ + tan$$70^\circ$$) is equal to

Solution

(1 + sec $$20^\circ$$ + cot $$70^\circ$$)(1 - cosec $$20^\circ$$ + tan$$70^\circ$$)

= (1 + sec $$20^\circ$$ + tan $$20^\circ$$)(1 - cosec $$20^\circ$$ + cot$$20^\circ$$)

= (1 + $$\frac{1}{cos20^\circ}$$ + $$\frac{sin20^\circ}{cos20^\circ}$$)(1 - $$\frac{1}{sin20^\circ}$$ + $$\frac{cos20^\circ}{sin20^\circ}$$)

= ($$\frac{1+cos20^\circ+sin20^\circ}{cos20^\circ}$$)($$\frac{sin20^\circ-1+cos20^\circ}{sin20^\circ}$$)

=  $$\frac{(cos20^\circ+sin20^\circ)^2-1}{cos20^\circ sin20^\circ}$$

= $$\frac{2cos20^\circ sin20^\circ}{cos20^\circ sin20^\circ}$$

=2


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