Question

(viii) tan(45° + a) tan(45º - a) = 1​

Answer 1

Answer:

tan(45° + a) + tan(45° - a)

=> \sf\dfrac{tan45 + tana}{1 - tan45.tana} + \dfrac{tan45 - tana}{1 + tan45tana}

=> \sf\dfrac{1 + tana}{1 - 1(tana)} + \dfrac{1 - tana}{1 + 1(tana)}

=> \sf\dfrac{1 + tana}{1 - tana} + \dfrac{1 - tana}{1 + tana}

=> \sf\dfrac{(1+tana)(1+tana) + (1-tana)(1-tana)}{(1+tana)(1-tana)}

=> \sf\dfrac{(1+tana)^{2} + (1-tana)^{2}}{1^{2} - tana^{2}}

=> \sf\dfrac{(1)^{2} + (tana)^{2} + 2tana + (1)^{2} + (tana)^{2} - 2tana}{1 - tan^{2}a}

=> \sf\dfrac{1 + tan^{2}a + 2tana + 1 + tan^{2}a - 2tana}{1 - tan^{2}a}

=> \sf\dfrac{2 + 2tan^{2}a}{1 - tan^{2}a}

=> \sf\dfrac{2(1 + tana)(1 + tana)}{(1 + tana)(1 - tana)}

=> \sf\dfrac{2 + 2tana}{1 - tana}

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