Question

Prove that: cot15degree+cot75degree=sec^215degree/(root over sec^215degree-1)

Answer 1

Step-by-step explanation:

Given:

{\star{\sf{ \: 4 {x}^{2} - 4x - 8 }}}⋆4x

2

−4x−8

{\bf{\blue{\underline{Now:}}}}

Now:

Divide by 4,

\begin{lgathered}{ : \implies{\sf{ \: {x}^{2} - x - 2 = 0 }}}\\ \\\end{lgathered}

:⟹x

2

−x−2=0

\begin{lgathered}{ : \implies{\sf{ \: {x}^{2} - 2x + x- 2 = 0 }}}\\ \\\end{lgathered}

:⟹x

2

−2x+x−2=0

\begin{lgathered}{ : \implies{\sf{ \: x(x - 2) - 1(x - 2) = 0 }}}\\ \\\end{lgathered}

:⟹x(x−2)−1(x−2)=0

\begin{lgathered}{ : \implies{\sf{ \: (x + 1)(x - 2) = 0 }}}\\ \\\end{lgathered}

:⟹(x+1)(x−2)=0

Take,

\begin{lgathered}{ : \implies{\sf{ \: (x + 1) = 0 \: \: \: \: \: \: and \: \: \: \: \: (x - 2) = 0 }}}\\ \\\end{lgathered}

:⟹(x+1)=0and(x−2)=0

\begin{lgathered}{ : \implies{ {\boxed{\sf{ \: x = - 1}} \: \: \: \: \: \: and \: \: \: \: \: { \boxed { \sf\: x = 2 }}}}}\\ \\\end{lgathered}

:⟹

x=−1

and

x=2

So, the value of 4x²-4x-8 is zero when x=-1,2

Therefore,the zeros of 4x²-4x-8 are -1 and 2.

Now,

\begin{lgathered}\star\boxed{\sf{ \purple {Sum \: of \: zeros \: = \frac{ - (Coefficient \: of \: x)}{Coefficient \: of \: {x}^{2} } }}}\\ \\\end{lgathered}

⋆

Sumofzeros=

Coefficientofx

2

−(Coefficientofx)

\begin{lgathered}{ : \implies{\sf{ - 1 + 2 = \frac{ - ( - 4)}{ 4} }}}\\ \\\end{lgathered}

:⟹−1+2=

4

−(−4)

\begin{lgathered}{ : \implies{\sf{ 1 = \frac{ - ( - 1)}{ 1} }}}\\ \\\end{lgathered}

:⟹1=

1

−(−1)

\begin{lgathered}{ : \implies \boxed{\sf{ 1 = 1 }}}\\ \\\end{lgathered}

:⟹

1=1

\begin{lgathered}\star\boxed{\sf{ \purple {Product \: of \: zeros \: = \frac{ constant \: term}{coefficient \: of \: {x}^{2} } }}}\\ \\\end{lgathered}

⋆

Productofzeros=

coefficientofx

2

constantterm

\begin{lgathered}{ : \implies {\sf{ ( - 1 )(2) = \frac{ - 8}{4} }}}\\ \\\end{lgathered}

:⟹(−1)(2)=

4

−8

: \implies\boxed {\sf{ - 2 = - 2 }}:⟹

−2=−2

Hence Verified.