Question

please solve this question it is an emergency please​

Answer 1

Question:

Prove that :

${sec}^{6} \theta = {tan}^{6} \theta + 3 {tan}^{2} \theta. {sec}^{2} \theta + 1$

Note:

• sin²@ + cos²@ = 1

• cosec²@ - cot²@ = 1

• sec²@ - tan²@ = 1

• (A+B)³ = A³ + 3•A•B•(A+B) + B³

Proof:

$= > lhs = {sec}^{6} \theta \\ = >lhs = { ({sec}^{2} \theta) }^{3} \\ = > lhs = { ({tan}^{2} \theta + 1) }^{3} \\ = > lhs = { ({tan}^{2} \theta) }^{3} + 3 . {tan}^{2} \theta . 1 . ( {tan}^{2} \theta + 1 ) + {1}^{3} \\ = > lhs = {tan}^{6} \theta + 3. {tan}^{2} \theta. {sec}^{2} \theta + 1 \\ = > lhs = rhs$

Hence proved.

Answer 2

$\Large\underline\mathfrak{Question}$

Prove :- sec^6@ = tan^6@ + 3tan²@*sec²@ + 1

$\Large\bold\star\underline{\underline\textsf{Formula\:used}}$

• a^(p*q) = (a^(p)*)^q
• sec²@ = (1+tan²@)
• (a+b)³ = a³ + b³ + 3ab(a+b)

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$\underline {\underline{\LARGE{{\bf{\green{S}}}{\mathfrak{o}}{\mathfrak{\orange{l}}}{\mathfrak{\red{u}}}{\mathfrak{\pink{t}}}{\mathfrak{\purple{i}}}{\mathfrak{\blue{o}}}{\mathfrak{\red{n}}}}}} : \:$

$\bf \: \green{Solving}, LHS , \: \\ \\ \red\longrightarrow \sf sec^{6} \theta = (sec^{2} \theta)^{3} \\ \\ \red\longrightarrow \sf \: (1 + {tan}^{2} \theta)^{3} \\ \\ \red\longrightarrow \sf 1^{3} +( {tan}^{2} \theta)^{3} + 3 \times 1 \times {tan}^{2} \theta(1 + {tan}^{2} \theta) \\ \\ \red\longrightarrow \bf \blue{ 1 + {tan}^{6} \theta \: + 3{tan}^{2} \theta \times sec^{2} \theta} \\ \\ \boxed{\red{\textbf{= RHS (Proved)}}} \:$

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$\orange{\huge\bf \: proof \: with} \: \huge\mathbb{RHS} \:$

$\bf\green{tan^{6}A+3tan^{2}A\:\times\:sec^{2}A+1} \\ \\=\sf tan^{6}A + 3tan^{2}A(1+ tan^{2}A)+1 \\ \\ \bf since \:\boxed{\sf \: sec^{2}A = 1+ Tan^{2}A} \\ \\\sf \: tan^{6} A + 3tan^{4}A + 3 tan^{2} A + 1 \\ = \sf \: (Tan^{2}A)^{3} + 3(tan^{2} A)^{2} + 3 (tan^{2} A) + 1 \\ \\\blue{\sf It \: is \: in \: the \: \: form \: of}\\ \\\purple{\boxed{\bf a^{3}+3a^{2} b+3ab^{2}+b^{3}=(a+b)^{3}}} \\ \\ \sf ∴(tan^{2} A)^{3}+3(tan^{2} A)^{2}+3(tan^{2}A)+1\\=\sf (tan^{2}A+1)^{3}\\=\sf (sec^{2} A)^{3}\\=\huge\red{\boxed{\bf \: sec^{6}A}}$

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#BAL

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