Solve tan(sec^-1 25/7)
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5
Let sec⁻¹ 25/7=α
then, secα=25/7=hypotenuse/base
By Pythagoras's theorem, perpendicular²+base²=hypotenuse²
or, perpendicular=√(hypotenuse²-base²)
or, perpendicular= √(25²-7²)=√(625-49)=√576=24
∴, tanα=perpendicular/base=24/7
Then, tan(sec⁻¹25/7)=tanα=24/7
then, secα=25/7=hypotenuse/base
By Pythagoras's theorem, perpendicular²+base²=hypotenuse²
or, perpendicular=√(hypotenuse²-base²)
or, perpendicular= √(25²-7²)=√(625-49)=√576=24
∴, tanα=perpendicular/base=24/7
Then, tan(sec⁻¹25/7)=tanα=24/7
Answered by
19
1 + tan2 θ = sec2 θ
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