Question

help me the question is given in attachment ...U have to prove it​

Answer 1

Answer:

see on LHS side

Step-by-step explanation:the numerator is in the form a^3-b^3 so the formula for a^3-b^3=(a-b)(a^2+ab+b^2)

so u will get

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

tan theta-1 will be canceled as its present on both numerator and denominator

u r left with (tan^2+1+tan theta)  (formula:tan^2+1=sec^2 theta)

so substitute

u will get sec^2 theta+tan theta

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