Question

sec⁴theta(1-sin⁴ theta)-2 tan⁴theta=1​

Answer 1

Step-by-step explanation:

Answer:-

$\sf{sec^4 \theta (1 - sin^4 \theta)-2 tan^4 \theta}$

Under the Property-

$\sf{a^2-b^2 = (a-b)(a+b)}$ [Or we may not be using it though]

Let's Do!

$\sf{sec^4 \theta (1 - sin^4 \theta)-2 tan^4 \theta}$

$\sf{sec^4 \theta ((1 - sin^2 \theta)(1 + sin^2 \theta))-2 tan^4 \theta}$

$\sf{sec^4 \theta (cos^2 \theta \times (1 + sin^2 \theta))-2 tan^4 \theta}$

As ,

$\sf{1-sin^2 \theta = cos^2 \theta}$

$\sf{sec^4 \theta \times cos^2 \theta \times (1 + sin^2 \theta)-2 tan^4 \theta}$

$\sf{sec^4 \theta \times \dfrac{1}{sec^2 \theta}\times (1 + sin^2 \theta)-2 tan^4 \theta}$

Cancelling them, we get

$\sf{sec^2 \theta (1 + sin^2 \theta)-2 tan^4 \theta}$

$\sf{sec^2 \theta+ \dfrac{sin^2 \theta}{cos^2 \theta})-2 tan^4 \theta}$

$\sf{sec^2 \theta+ tan^2 \theta -2 tan^4 \theta}$

Now, sec² theta - tan² theta = 1

So, cancelling it,

$\sf{sec^2 \theta - tan^2 \theta = 1}$

Hence Proved!

Answer 2

Given

sec⁴theta(1-sin⁴ theta)-2 tan⁴theta=1

We Used

We used this Property -

a² - b² = (a - b) (a + b)

According to the question

$\sf{sec^4 \theta ((1 - sin^2 \theta)(1 + sin^2 \theta))-2 tan^4 \theta} \\ \\ </p><p></p><p>\sf{sec^4 \theta (cos^2 \theta \times (1 + sin^2 \theta))-2 tan^4 \theta} \\ \\ </p><p></p><p>As , \\ \\ </p><p></p><p>\sf{1-sin^2 \theta = cos^2 \theta} \\ \\ </p><p></p><p>\sf{sec^4 \theta \times cos^2 \theta \times (1 + sin^2 \theta)-2 tan^4 \theta} \\ \\ </p><p></p><p>\sf{sec^4 \theta \times \dfrac{1}{sec^2 \theta}\times (1 + sin^2 \theta)-2 tan^4 \theta} \\ \\ </p><p></p><p>Cancelling \:them, \:we\: get \\ \\ </p><p></p><p>\sf{sec^2 \theta (1 + sin^2 \theta)-2 tan^4 \theta} \\ \\ </p><p></p><p>\sf{sec^2 \theta+ \dfrac{sin^2 \theta}{cos^2 \theta})-2 tan^4 \theta} \\ \\ </p><p></p><p>\sf{sec^2 \theta+ tan^2 \theta -2 tan^4 \theta} \\ \\ </p><p></p><p>Now, sec² theta - tan² theta = 1 \\ \\ </p><p></p><p>So, \:cancelling \:it, \\ \\ </p><p></p><p>\sf{sec^2 \theta - tan^2 \theta = 1}$

Hence its verified