Math, asked by crazypie52, 1 month ago

if m= tan Theta + sin theta n=tan theta - sin theta prove that (m²-n²)²=16mn
Please answer in good content and full explanation Boad exam based question.​​​

Answers

Answered by MiraculousBabe
179

Answer:

\tt\small\underline\green{Given:-}

\tt{\implies m=tan\theta+sin\theta}

\tt{\implies n=tan\theta-sin\theta}

\tt\small\underline\green{To\:Prove:-}

\tt{\implies (m^2-n^2)^2=16mn}

\tt\small\underline\green{Solution:-}

Here we will simply both sides L.H.S and R.H.S by putting the value of m and n.

\tt\small\underline\green{Calculation\:for\:L.H.S:-}

\tt{\implies (m^2-n^2)^2}

\tt{\implies [(tan\theta+sin\theta)^2\:-(tan\theta-sin\theta)^2]^2}

\tt{\implies Here\: using\: identity}

\tt{\implies (a+b)^2=a^2+b^2+2ab}

\tt{\implies (a-b)^2=a^2+b^2-2ab}

\tt{\implies [(tan^2\theta+sin^2\theta+2tan\theta.sin\theta)-(tan^2\theta+sin^2\theta-2tan\theta.sin\theta)]^2}

\tt{\implies [\cancel{tan^2\theta}+\cancel{sin^2\theta}+2tan\theta.sin\theta-\cancel{tan^2\theta}-\cancel{sin^2\theta}+2tan\theta.sin\theta]^2}

\tt{\implies [2tan\theta.sin\theta+2tan\theta.sin\theta]^2}

\tt{\implies [4tan\theta.sin\theta]^2}

\tt{\implies 16tan^2\theta.sin^2\theta}

\tt\small\underline\green{Calculation\:for\:R.H.S:-}

\tt{\implies 16mn}

\tt{\implies 16[(tan\theta+sin\theta)(tan\theta-sin\theta)]\:}

\tt{\implies Here\: using\: identity}

\tt{\implies (a+b)(a-b)=a^2-b^2}

\tt{\implies 16[tan^2\theta-sin^2\theta]}

\tt{\implies 16[\dfrac{sin^2\theta}{cos^2\theta}-sin^2\theta]}

\tt{\implies 16[\dfrac{sin^2\theta-sin^2\theta.cos^2\theta}{cos^2\theta}]}

\tt{\implies 16.sin^2\theta[\dfrac{(1-cos^2\theta)}{cos^2\theta}]}

\tt{\implies \dfrac{16.sin^2\theta.sin^2\theta}{cos^2\theta}}

\tt{\implies \dfrac{16.sin^2\theta}{cos^2\theta}.sin^2\theta}

\tt{\implies 16tan^2\theta.sin^2\theta}

\sf\large{Hence\:R.H.S=L.HS}

Answered by Anonymous
5

Tan θ + sinθ = m Squaring on both sides& Tan θ - Sin θ = n Squaring on both sides we get

prove that ( - )²=16mn

Tan² θ + sin² θ + 2 Tan θ sinθ = ---------(1)

Tan² θ + sin² θ - 2 Tan θ sinθ = --------(2)

Substract (2) from (1) we get,

- = 4 Tanθ sinθ = 4 √( Tan²θ sin²θ)

= 4 √( Sin²θ /Cos²θ ( 1 - Cos²θ) )

= 4 √( Tan²θ - sin²θ)

= 4 √( Tan θ + Sin θ)(Tan θ - Sin θ )

= m2 - n2 = 4√mn ( on squaring both sides)

= (m2-n2)²=16mn

Step-by-step explanation:

hope it's helpful,,

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