Prove the following identity. Kindly, don't spam. Irrelevant answer will be reported.
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Prove the following identity. Kindly, don't spam. Irrelevant answer will be reported.
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To prove:
[tex]\sf{(cosec \ A-sin \ A)(sec \ A-cos \ A) \ sec^{2}A=tan \ A}[/tex]
Proof:
[tex]\sf{L.H.S.=(cosec \ A-sin \ A)(sec \ A-cos \ A) \ sec^{2}A}[/tex]
[tex]\sf\blue{But, \ cosec\theta=\dfrac{1}{sin\theta} \ and \ sec\theta=\dfrac{1}{cos\theta}}[/tex]
[tex]\sf{=(\dfrac{1}{sin \ A}-sin \ A)(\dfrac{1}{cos \ A}-cos \ A) \ sec^{2}A}[/tex]
[tex]\sf{=(\dfrac{1-sin^{2}A}{sin \ A})(\dfrac{1-cos^{2}A}{cos \ A}) \ sec^{2}A}[/tex]
[tex]\sf\blue{1-sin^{2}\theta=cos^{2}\theta \ and \ 1-cos^{2}\theta=sin^{2}\theta}[/tex]
[tex]\sf{=(\dfrac{cos^{2}A}{sin \ A})(\dfrac{sin^{2}A}{cos \ A}) \ sec \ A}[/tex]
[tex]\sf{=(cos \ A.sin \ A) \ sec^{2}A}[/tex]
[tex]\sf{=(cos \ A.sin \ A) \dfrac{1}{cos^{2}A}}[/tex]
[tex]\sf{=\dfrac{sin \ A}{cos \ A}}[/tex]
[tex]\sf{=tan \ A}[/tex]
[tex]\sf{=R.H.S.}[/tex]
[tex]\sf{Hence, \ proved.}[/tex]
[tex]\sf\purple{\tt{(cosec \ A-sin \ A)(sec \ A-cos \ A) \ sec^{2}A=tan \ A}}[/tex]
Step-by-step explanation:
L.H.S. = (cosec A – sin A)(sec A – cos A). sec² A
= (1/sin A – sin A). (1/cos A – cos A). 1/cos² A
= (1 – sin² A/sin A) × (1 – cos2 A/cos A) × 1/cos2 a
= cos² A/sin A × sin² A/cos A × 1/cos² A
[∵ (1 – sin² A) = cos² A]
[∵ 1 – cos² A = sin²A]
= sin A/cos A = tan A
= R.H.S.
Proved (cosec A – sin A) (sec A – cos A) sec² A = tan A.