Question

I need help!! I will give 5.0 star ​

Answer 1

So, here we need to differentiate tan³$(\theta)$, wr.t.$\theta$, but let's recall some identities which will be very useful in this question :

• ${\boxed{\bf{\dfrac{d}{dx}\{\tan (x)\}=\sec^{2}(x)}}}$

• ${\boxed{\bf{\dfrac{d}{dx}(x^n)=nx^{n-1}}}}$

• ${\boxed{\bf{\sec^{2}(\theta)=\tan^{2}(\theta)+1}}}$

Coming back on the question, consider :

${:\implies \quad \sf \dfrac{d}{d\theta}\{\tan^{3}(\theta)\}}$

${:\implies \quad \sf 3\tan^{3-1}(\theta)\dfrac{d}{d\theta}\{\tan (\theta)\}}$

${:\implies \quad \sf 3\tan^{2}(\theta)\sec^{2}(\theta)}$

Using the identity ;

${:\implies \quad \sf 3\tan^{2}(\theta)\{1+\tan^{2}(\theta)\}}$

${:\implies \quad \sf 3\tan^{2}(\theta)+3tan^{4}(\theta)}$

${:\implies \quad \sf 3\tan^{4}(\theta)+3\{\sec^{2}(\theta)-1\}}$

${:\implies \quad \boxed{\bf 3\tan^{4}(\theta)+3\sec^{2}(\theta)-3}}$

Hence, Proved