Only for Brainly moderators , Brainly stars and brainly best users !▬▬ஜ۩۞۩ஜ▬▬ 1 + cot A + tan a into Sin A minus Cos A is equal to sec a upon cos square A minus Cos A upon sec square A.. Solve on paper
Answers
Answer:
To prove: (1+\cot A +\tan A)(\sin A -\cos A)=\dfrac{\sec A}{\text {cosec}^2 A}- \dfrac{\text {cosec}}{\sec ^2 A}(1+cotA+tanA)(sinA−cosA)=
cosec
2
A
secA
−
sec
2
A
cosec
Step-by-step explanation:
Consider L.H.S.
\begin{gathered}(1+\cot A +\tan A)(\sin A -\cos A)\\\\= (1 + \dfrac{\cos A}{\sin A}+\dfrac{\sin A}{\cos A})(\sin A- \cos A)\\\\= (1+ \dfrac{\cos^2 A +\sin ^2 A}{\sin A \cos A} )(\sin A- \cos A)\\\\=(1+\dfrac{1}{\sin A \cos A}) (\sin A- \cos A) \tex{------ }(\because \sin^2A +\cos^2A =1})\\\\= \dfrac{1+\sin A \cos A}{\sin A \cos A} (\sin A- \cos A)\end{gathered}
Now consider R.H.S.
we have
\begin{gathered}\dfrac{\sec A}{\text {cosec}^2 A}- \dfrac{\text {cosec}}{\sec ^2 A}\\\\=\dfrac{\sin^2 A}{\cos A} -\dfrac{\cos^2 A}{\sin A} \\\\= \dfrac{\sin^3 A-\cos ^3 A}{\sin A\cos A} \\\\= \dfrac{(\sin A-\cos A)(\sin ^2 A +\cos ^2 A + \sin A\cos A)}{\sin A \cos A} \\\\\text{}[\because a^3-b^3=(a-b)(a^2+b^2+ab)]\\\\=\dfrac{(\sin A-\cos A)(1 + \sin A\cos A)}{\sin A \cos A}\end{gathered}
cosec
2
A
secA
−
sec
2
A
cosec
=
cosA
sin
2
A
−
sinA
cos
2
A
=
sinAcosA
sin
3
A−cos
3
A
=
sinAcosA
(sinA−cosA)(sin
2
A+cos
2
A+sinAcosA)
[∵a
3
−b
3
=(a−b)(a
2
+b
2
+ab)]=
sinAcosA
(sinA−cosA)(1+sinAcosA)
Now as L.H.S. = R.H.S
Hence, proved the required result
Step-by-step explanation:
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