Question

prove that LHS equals RHS
please help me only a mastermind can solv it it's my challenge ​

Answer 1

Answer:

See below

Step-by-step explanation:

We'll start with the left-hand side (LHS) and manipulate it to arrive at the right-hand side (RHS).

LHS:

$\sf \sec^2(\theta) \left(1 - \dfrac{\sin^2(\theta)}{\sin^2(A)}\right)$

Express $\sf \sec^2(\theta)$ in terms of $\sf \sin^2(\theta)$ and $\sf \cos^2(\theta)$:

$\sf = \dfrac{1}{\cos^2(\theta)} \left(1 - \dfrac{\sin^2(\theta)}{\sin^2(A)}\right)$

Distribute the $\sf \dfrac{1}{\cos^2(\theta)}$ term:

$\sf = \dfrac{1}{\cos^2(\theta)} - \dfrac{\sin^2(\theta)}{\cos^2(\theta)\sin^2(A)}$

Rewrite $\sf \dfrac{\sin^2(\theta)}{\cos^2(\theta)}$ as $\sf \tan^2(\theta)$:

$\sf = \dfrac{1}{\cos^2(\theta)} - \dfrac{\tan^2(\theta)}{\sin^2(A)}$

Express $\sf \dfrac{1}{\cos^2(\theta)}$ as $\sf 1 + \tan^2(\theta)$:

Here is how it becomes: we know that:

$\sf sin^2 \theta + cos^2 \theta = 1$

using this,

$\boxed{\boxed{\begin{aligned} \dfrac{1}{cos^2 \theta } & = \dfrac{sin^2 \theta + cos^2 \theta}{ cos^2 \theta } \\\\ & = \dfrac{sin^2\theta }{cos^2 \theta } + \dfrac{cos^2\theta }{cos^2 \theta } \\\\ & = tan^2\theta + 1 \\\\ & = 1+ tan^2\theta \end{aligned}}}$

Replace the value of $\sf \dfrac{1}{cos^2 \theta }$

$\sf = \left(1 + \tan^2(\theta)\right) - \dfrac{\tan^2(\theta)}{\sin^2(A)}$

Combine like terms:

$\sf = 1 - \tan^2(\theta) \left(1/\sin^2(A) - 1\right)$

Express $\sf 1/\sin^2(A)$ as $\sf \csc^2(A)$:

$\sf = 1 - \tan^2(\theta)\left(\csc^2(A) - 1\right)$

Recognize that $\sf \csc^2(A) - 1 = \cot^2(A)$:

Here is how it is:

$\boxed{\boxed{\begin{aligned} csc ^2 A - 1 & = \dfrac{1}{sin^2 A} - 1 \\\\ & = \dfrac{ 1- sin^2 A }{ sin^2 A}\\\\ &= \dfrac{ (sin^2 A + cos^2 A) - sin^2 A}{sin^2 A} \\\\ & = \dfrac{cos^2 A}{sin^2 A}\\\\ & = cot^2 A \end{aligned}}}$

Now, replace the value of $\sf csc ^2 A - 1$

$\sf = 1 - \tan^2(\theta)\cot^2(A)$

Express $\sf \cot^2(A)$ as $\sf 1/\tan^2(A)$:

$\sf = 1 - \dfrac{\tan^2(\theta)}{\tan^2(A)}$

This matches the right-hand side (RHS), so the identity is proven.

Therefore:

$\sf \sec^2(\theta) \left(1 - \dfrac{\sin^2(\theta)}{\sin^2(A)}\right) = 1 - \dfrac{\tan^2(\theta)}{\tan^2(A)}$