Question

If sec A = x + 1/4x , then sec A + tan A equals what ❓

Answer 1

Step-by-step explanation:

Givensecθ=x+4x1−−(1)

/* We know the, trigonometric identity :

tan²A = Sec²A-1*/

\begin{lgathered}tan^{2}\theta \\= \left(x+\frac{1}{4x}\right)^{2}-1\\=x^{2}+2\times x \times \big(\frac{1}{4x}\big)+\left(\frac{1}{4x}\right)^{2}-1\end{lgathered}tan2θ=(x+4x1)2−1=x2+2×x×(4x1)+(4x1)2−1

=x^{2}+\frac{1}{2}+\frac{1}{16x^{2}}-1=x2+21+16x21−1

=x^{2}-\frac{1}{2}+\frac{1}{16x^{2}}=x2−21+16x21

=x^{2}-2\times x \times \frac{1}{4x}+\big(\frac{1}{4x}\big)^{2}=x2−2×x×4x1+(4x1)2

=\left(x-\frac{1}{4x}\right)^{2}=(x−4x1)2

tan\theta = ± \left(x-\frac{1}{4x}\right)--(2)tanθ=±(x−4x1)−−(2)

\begin{lgathered}Now,\\sec\theta + tan\theta\\=x+\frac{1}{4x}±\left(x-\frac{1}{4x}\right)\end{lgathered}Now,secθ+tanθ=x+4x1±(x−4x1)

\begin{lgathered}Case \:1\\sec\theta + tan\theta\\=x+\frac{1}{4x}+\left(x-\frac{1}{4x}\right)\end{lgathered}Case1secθ+tanθ=x+4x1+(x−4x1)

=2x=2x

\begin{lgathered}Case\:2\\sec\theta + tan\theta\\=x+\frac{1}{4x}-\left(x-\frac{1}{4x}\right)\end{lgathered}Case2secθ+tanθ=x+4x1−(x−4x1)

=x+\frac{1}{4x}-x+\frac{1}{4x}=x+4x1−x+4x1

=\frac{1}{2x}=

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