Find HCF by common division methodof 16,42,72​

Answer:

1.-d.

2.-e.

3.-b.

4.-c.

5.-a.

Answer:

1 will be with d

2 will be with e

3 will be with b

4 will be with c

5 will be with a

I hope this helps

Explanation:

Nothing to explain

Answer:

here is your answer

Step-by-step explanation:

(x+y)(x² - xy + y²)

= x ( x² - xy + y² ) +y ( x² - xy + y²)

= x³ - x²y + xy² +yx² -xy² + y³

= x³ + y³

Answer:

x^3-x^2y+xy^2+y^3

Step-by-step explanation:

thank you you

Step-by-step explanation:

Consider,

[tex]\qquad\sf \: {tan}^{2}A - {sin}^{2}A \\ \\ [/tex]

can be rewritten as

[tex]\qquad\sf \: = \: \dfrac{ {sin}^{2} A}{ {cos}^{2} A} - {sin}^{2}A \\ \\ [/tex]

[tex]\qquad\sf \: = \:\left( \dfrac{1}{ {cos}^{2} A} - 1\right) {sin}^{2}A \\ \\ [/tex]

[tex]\qquad\sf \: = \:\left( {sec}^{2}A - 1\right) {sin}^{2}A \\ \\ [/tex]

[tex]\qquad\sf \: = \: ({tan}^{2}A ) {sin}^{2}A \\ \\ [/tex]

[tex]\qquad\sf \: = \: {tan}^{2}A \: {sin}^{2}A \\ \\ [/tex]

Hence,

[tex]\qquad\sf\implies \bf \: {tan}^{2}A - {sin}^{2}A = {tan}^{2}A \: {sin}^{2}A \\ \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Formulae Used

[tex]\sf \:tanx \: = \: \dfrac{sinx}{cosx} \\ \\ [/tex]

[tex]\sf \:secx \: = \: \dfrac{1}{cosx} \\ \\ [/tex]

[tex]\sf \: {sec}^{2}x - {tan}^{2}x = 1 \\ \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Additional Information

[tex]\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: Formulae}}}} \\ \\ \bigstar \: \bf{sinx = \dfrac{1}{cosecx} }\\ \\ \bigstar \: \bf{cosx = \dfrac{1}{secx} }\\ \\ \bigstar \: \bf{tanx = \dfrac{sinx}{cosx} = \dfrac{1}{cotx} }\\ \\ \bigstar \: \bf{cot x= \dfrac{cosx}{sinx} = \dfrac{1}{tanx} }\\ \\ \bigstar \: \bf{cosec x) = \dfrac{1}{sinx} }\\ \\ \bigstar \: \bf{secx = \dfrac{1}{cosx} }\\ \\ \bigstar \: \bf{ {sin}^{2}x + {cos}^{2}x = 1 } \\ \\ \bigstar \: \bf{ {sec}^{2}x - {tan}^{2}x = 1 }\\ \\ \bigstar \: \bf{ {cosec}^{2}x - {cot}^{2}x = 1 } \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}[/tex]