12. Prove the following identity: (cosec A - sin A) (sec A - cos A) sec- A = tan A.
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Answer:
=(
sinA
1
−sinA)(
cosA
1
−cosA)(
cosA
sinA
+
sinA
cosA
)
= (\dfrac{1-sin^{2}A}{sinA})(\dfrac{1-cos^{2}A}{cosA})(\dfrac{sin^{2}A + cos^{2}A}{sinA\:cosA}=(
sinA
1−sin
2
A
)(
cosA
1−cos
2
A
)(
sinAcosA
sin
2
A+cos
2
A
= (\dfrac{cos^{2}A}{sinA})(\dfrac{sin^{2}A}{cosA})(\dfrac{1}{sinA\:cosA})=(
sinA
cos
2
A
)(
cosA
sin
2
A
)(
sinAcosA
1
)
= \dfrac{cosA \: sinA}{sinA \: cosA} = 1=
sinAcosA
cosAsinA
=1
Hence, proved.
☆Identities used☆
• cosec\theta = \dfrac{1}{sin\theta}cosecθ=
sinθ
1
• sec\theta = \dfrac{1}{cos\theta}secθ=
cosθ
1
• sin^{2}\theta + cos^{2}\theta = 1sin
2
θ+cos
2
θ=1
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