sec0 +tan0 = 1/3 find sec 0 - tan0
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Answered by
1
sec^20-tan^20=1;
(seco+tan0)(sec0-tan0)=1;
1/3(sec0-tan0)=1;
sec0-tan0=3;
(seco+tan0)(sec0-tan0)=1;
1/3(sec0-tan0)=1;
sec0-tan0=3;
Answered by
2
sec²θ-tan²θ= 1
(secθ+tanθ)(secθ-tanθ)=1
given,
secθ+tanθ=1/3
so, secθ-tanθ=3
(secθ+tanθ)(secθ-tanθ)=1
given,
secθ+tanθ=1/3
so, secθ-tanθ=3
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