cotB=512, prove that: tan2B−sin2B=sin2Bsec2B

Question

cotB=512​, prove that: tan2B−sin2B=sin2Bsec2B

Filo · Accepted Answer

we know thatcotθ =sideoppositetoangleθ sideadjacenttoangleθ ​orcotθ =PerpendicularBase​cosB=512​⇒PB​=512​⇒BCAB​=512​letside adjacent to angle B=AB=12kside opposite to angle B=BC=5kwhere k is any positive integerfirstly we have to find the value of AC using Pythagoras theorem⇒(AB) 2+(BC) 2=(AC) 2⇒(12k) 2+(5k) 2=(AC) 2⇒(AC) 2=144k2+25k2⇒(AC) 2=169k2⇒AC=±13kbut side AC cannot be negative so AC=13kwe know thatsinθ =hypotenusesideoppositetoangleθ ​side opposite to angle B=BC=5khypotenuse=AC=13kso,sinB=ACBC​=13k5k​=135​we know thatcosθ =hypotenusesideadjacenttoangleθ ​side adjacent to angle B=AB=12khypotenuse =AC=13kso, cosB=ACAB​=13k12k​=1312​tanB=BP​=ABBC​=5k12k​=125​;secB=cosB 1​=1312​ 1​=1213​Now,LHS=tan2B−sin2B=(125​ ) 2−(135​ ) 2=14425​−16925​=24336625​RHS=sin4Bsec2B=(135​ ) 4−(1213​ ) 2=(135​ ) 2(135​ ) 2(1213​ ) 2=24336625​LHS=RHS

Question Text	$cot B = \frac{12}{5}$ , prove that: $tan^{2} B - sin^{2} B = sin^{2} B sec^{2} B$
Answer Type	Text solution:1
Upvotes	150

$cot B = \frac{12}{5}$ , prove that: $tan^{2} B - sin^{2} B = sin^{2} B sec^{2} B$

Text solutionVerified

Practice questions from similar books

cotB=512​, prove that: tan2B−sin2B=sin2Bsec2B

Text solutionVerified

Practice questions from similar books

$cot B = \frac{12}{5}$ , prove that: $tan^{2} B - sin^{2} B = sin^{2} B sec^{2} B$