Question

sec^4θ-sec^2θ=tan^4θ+tan^2θ

Answer 1

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sec^2a= 1+tan^a

now LHS

sec^4-sec^2

(sec^2)^2- sec^2

(1+tan^2)^2-(1+tan^2)

1+tan^4+2tan^2-1-tan^2

tan^4+tan^2

Answer 2

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

$\rightarrow\boxed{tan^{4}\theta+tan^{2}\theta}$

$\rightarrow\boxed{tan^{2}\theta(tan^{2}\theta+1}$

${\boxed{Using\rightarrow1+tan^{2}\theta =sec^{2}\theta}}$

${\boxed{Using\rightarrow tan^{2}\theta =sec^{2}\theta-1}}$

$\rightarrow\boxed{tan^{2}\theta(sec^{2}\theta}$

$\rightarrow\boxed{(sec^{2}\theta-1)sec^{2}\theta}$

$\rightarrow\boxed{sec^{4}\theta-sec^{2}\theta}$

★RHS = LHS

$\boxed{\begin{minipage}{7 cm} Fundamental Trignometric Indentities \\ \\ \tan (90 - A) = cotA \\ \\ cot (90 - A) = tanA \\ \\ sec (90 - A) = cosecA \\ \\ tan\theta =\dfrac{sin\theta}{cos\theta} \\ \\ cot\theta =\dfrac{cos\theta}{sin\theta} \\ \\ cosec (90 - A) = secA \\ \\ sin^{2}\theta+\cos^{2}\theta =1\\ \\ 1+tan^{2}\theta=\sec^{2}\theta \\ \\ 1 + cot^{2}\theta=\text{cosec}^2\theta \end{minipage}}$